SQLSTATE[42000]: Syntax error or access violation: 1064 Error in setting Variable in bindparam











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I want to get data from DB and add string to theme and put them in another query . When I run code I've got this error :




Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or
access violation: 1064 You have an error in your SQL syntax; check the
manual that corresponds to your MySQL server version for the right
syntax to use near ''goldhyipPID', 'goldhyipPayStatus',
programName,'goldhyipLastPayout') VALUES ('1' at line 1 in
C:wamp64wwwallmonitorstest.php on line 69




<?php
set_time_limit(3600);
require_once 'fetchdetails/func.php';
require_once 'config.php';
$stmt = $conn->prepare("SELECT * FROM `monitors`");
$stmt->execute();
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
foreach ($result as $monitor) {
$monitorName = $monitor['monitorName'];
$monitorNamePID = $monitorName . 'PID';
$monitorNameLastPayout = $monitorName . 'LastPayout';
$monitorNamePayStatus = $monitorName . 'PayStatus';
$siteURL = $monitor['monitorurl'];
$pattern1GetPID = $monitor['monitorPatternGetPID'];
$patternLastPayOut = $monitor['monitorPatternLastPayout'];
$patternPStatus = $monitor['monitorPatternPayStatus'];
$patterndetailsurl = $monitor['monitorDetailsLink'];
$patterngotositesurl = $monitor['monitorPatternGoSite'];
$content = getPageContent($siteURL);
preg_match_all($pattern1GetPID, $content, $matches, PREG_SET_ORDER, 0);
foreach ($matches as $pid) {
$id = $pid[1];
$detailsurl = $patterndetailsurl . $id;
$gositesurl = $patterngotositesurl . $id;
$details = getPageContent($detailsurl);
preg_match_all($patternPStatus, $details, $status);
$payingStatusNumber = $status[1][0];
if ($payingStatusNumber == 4) {
$payingStatus = 'Not Paying';
} elseif ($payingStatusNumber == 3) {
$payingStatus = 'Problem';
} elseif ($payingStatusNumber == 2) {
$payingStatus = 'Waiting';
} elseif ($payingStatusNumber == 1) {
$payingStatus = 'Paying';
}
preg_match_all($patternLastPayOut, $details, $payout);
if (isset($payout[1][0])) {
$payoutdate = $payout[1][0];
} else {
$payoutdate = ' Not Set';
};
$stmt2 = $conn->prepare('SELECT * FROM programs where :monitorNamePID=:id');
$stmt2->bindParam('monitorNamePID', $monitorNamePID);
$stmt2->bindParam('id', $id);
$stmt2->execute();
$numofupdates = $stmt2->rowCount();
if ($numofupdates >= 1) {
$stmt3 = $conn->prepare("UPDATE programs SET :monitorNamePayStatus=:payingstatus , :monitorNameLastPayout=:goldhyiplastpayout WHERE :monitorNamePID=:goldhyippid ");
$stmt3->bindParam('monitorNamePayStatus', $monitorNamePayStatus);
$stmt3->bindParam('monitorNameLastPayout', $monitorNameLastPayout);
$stmt3->bindParam('monitorNamePID', $monitorNamePID);
$stmt3->bindParam('payingstatus', $payingStatus);
$stmt3->bindParam('goldhyiplastpayout', $payoutdate);
$stmt3->bindParam('goldhyippid', $id);
$stmt3->execute();
echo 'P ID Updated : ' . $id . '<br>';
} else {
$siteAddress = get_redirect_final_host_url($gositesurl);
echo $siteAddress;
$stmt3 = $conn->prepare('INSERT INTO programs (:monitorNamePID, :monitorNamePayStatus, programName,:monitorNameLastPayout) VALUES (:goldhyippid, :payingstatus, :progname, :goldhyiplastpayout)');
$stmt3->bindParam('monitorNamePID', $monitorNamePID);
$stmt3->bindParam('monitorNamePayStatus', $monitorNamePayStatus);
$stmt3->bindParam('monitorNameLastPayout', $monitorNameLastPayout);
$stmt3->bindParam('goldhyippid', $id);
$stmt3->bindParam('payingstatus', $payingStatus);
$stmt3->bindParam('progname', $siteAddress);
$stmt3->bindParam('goldhyiplastpayout', $payoutdate);
$stmt3->execute();
echo 'P ID inserted : ' . $id . '<br>';
}
}
echo "Fetching $siteURL Done <br>";
}


I took details from sql, and add string to them, use in PDO query. When I write column name its ok but when I use Variable I got error.










share|improve this question




























    up vote
    0
    down vote

    favorite












    I want to get data from DB and add string to theme and put them in another query . When I run code I've got this error :




    Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or
    access violation: 1064 You have an error in your SQL syntax; check the
    manual that corresponds to your MySQL server version for the right
    syntax to use near ''goldhyipPID', 'goldhyipPayStatus',
    programName,'goldhyipLastPayout') VALUES ('1' at line 1 in
    C:wamp64wwwallmonitorstest.php on line 69




    <?php
    set_time_limit(3600);
    require_once 'fetchdetails/func.php';
    require_once 'config.php';
    $stmt = $conn->prepare("SELECT * FROM `monitors`");
    $stmt->execute();
    $result = $stmt->fetchAll(PDO::FETCH_ASSOC);
    foreach ($result as $monitor) {
    $monitorName = $monitor['monitorName'];
    $monitorNamePID = $monitorName . 'PID';
    $monitorNameLastPayout = $monitorName . 'LastPayout';
    $monitorNamePayStatus = $monitorName . 'PayStatus';
    $siteURL = $monitor['monitorurl'];
    $pattern1GetPID = $monitor['monitorPatternGetPID'];
    $patternLastPayOut = $monitor['monitorPatternLastPayout'];
    $patternPStatus = $monitor['monitorPatternPayStatus'];
    $patterndetailsurl = $monitor['monitorDetailsLink'];
    $patterngotositesurl = $monitor['monitorPatternGoSite'];
    $content = getPageContent($siteURL);
    preg_match_all($pattern1GetPID, $content, $matches, PREG_SET_ORDER, 0);
    foreach ($matches as $pid) {
    $id = $pid[1];
    $detailsurl = $patterndetailsurl . $id;
    $gositesurl = $patterngotositesurl . $id;
    $details = getPageContent($detailsurl);
    preg_match_all($patternPStatus, $details, $status);
    $payingStatusNumber = $status[1][0];
    if ($payingStatusNumber == 4) {
    $payingStatus = 'Not Paying';
    } elseif ($payingStatusNumber == 3) {
    $payingStatus = 'Problem';
    } elseif ($payingStatusNumber == 2) {
    $payingStatus = 'Waiting';
    } elseif ($payingStatusNumber == 1) {
    $payingStatus = 'Paying';
    }
    preg_match_all($patternLastPayOut, $details, $payout);
    if (isset($payout[1][0])) {
    $payoutdate = $payout[1][0];
    } else {
    $payoutdate = ' Not Set';
    };
    $stmt2 = $conn->prepare('SELECT * FROM programs where :monitorNamePID=:id');
    $stmt2->bindParam('monitorNamePID', $monitorNamePID);
    $stmt2->bindParam('id', $id);
    $stmt2->execute();
    $numofupdates = $stmt2->rowCount();
    if ($numofupdates >= 1) {
    $stmt3 = $conn->prepare("UPDATE programs SET :monitorNamePayStatus=:payingstatus , :monitorNameLastPayout=:goldhyiplastpayout WHERE :monitorNamePID=:goldhyippid ");
    $stmt3->bindParam('monitorNamePayStatus', $monitorNamePayStatus);
    $stmt3->bindParam('monitorNameLastPayout', $monitorNameLastPayout);
    $stmt3->bindParam('monitorNamePID', $monitorNamePID);
    $stmt3->bindParam('payingstatus', $payingStatus);
    $stmt3->bindParam('goldhyiplastpayout', $payoutdate);
    $stmt3->bindParam('goldhyippid', $id);
    $stmt3->execute();
    echo 'P ID Updated : ' . $id . '<br>';
    } else {
    $siteAddress = get_redirect_final_host_url($gositesurl);
    echo $siteAddress;
    $stmt3 = $conn->prepare('INSERT INTO programs (:monitorNamePID, :monitorNamePayStatus, programName,:monitorNameLastPayout) VALUES (:goldhyippid, :payingstatus, :progname, :goldhyiplastpayout)');
    $stmt3->bindParam('monitorNamePID', $monitorNamePID);
    $stmt3->bindParam('monitorNamePayStatus', $monitorNamePayStatus);
    $stmt3->bindParam('monitorNameLastPayout', $monitorNameLastPayout);
    $stmt3->bindParam('goldhyippid', $id);
    $stmt3->bindParam('payingstatus', $payingStatus);
    $stmt3->bindParam('progname', $siteAddress);
    $stmt3->bindParam('goldhyiplastpayout', $payoutdate);
    $stmt3->execute();
    echo 'P ID inserted : ' . $id . '<br>';
    }
    }
    echo "Fetching $siteURL Done <br>";
    }


    I took details from sql, and add string to them, use in PDO query. When I write column name its ok but when I use Variable I got error.










    share|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I want to get data from DB and add string to theme and put them in another query . When I run code I've got this error :




      Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or
      access violation: 1064 You have an error in your SQL syntax; check the
      manual that corresponds to your MySQL server version for the right
      syntax to use near ''goldhyipPID', 'goldhyipPayStatus',
      programName,'goldhyipLastPayout') VALUES ('1' at line 1 in
      C:wamp64wwwallmonitorstest.php on line 69




      <?php
      set_time_limit(3600);
      require_once 'fetchdetails/func.php';
      require_once 'config.php';
      $stmt = $conn->prepare("SELECT * FROM `monitors`");
      $stmt->execute();
      $result = $stmt->fetchAll(PDO::FETCH_ASSOC);
      foreach ($result as $monitor) {
      $monitorName = $monitor['monitorName'];
      $monitorNamePID = $monitorName . 'PID';
      $monitorNameLastPayout = $monitorName . 'LastPayout';
      $monitorNamePayStatus = $monitorName . 'PayStatus';
      $siteURL = $monitor['monitorurl'];
      $pattern1GetPID = $monitor['monitorPatternGetPID'];
      $patternLastPayOut = $monitor['monitorPatternLastPayout'];
      $patternPStatus = $monitor['monitorPatternPayStatus'];
      $patterndetailsurl = $monitor['monitorDetailsLink'];
      $patterngotositesurl = $monitor['monitorPatternGoSite'];
      $content = getPageContent($siteURL);
      preg_match_all($pattern1GetPID, $content, $matches, PREG_SET_ORDER, 0);
      foreach ($matches as $pid) {
      $id = $pid[1];
      $detailsurl = $patterndetailsurl . $id;
      $gositesurl = $patterngotositesurl . $id;
      $details = getPageContent($detailsurl);
      preg_match_all($patternPStatus, $details, $status);
      $payingStatusNumber = $status[1][0];
      if ($payingStatusNumber == 4) {
      $payingStatus = 'Not Paying';
      } elseif ($payingStatusNumber == 3) {
      $payingStatus = 'Problem';
      } elseif ($payingStatusNumber == 2) {
      $payingStatus = 'Waiting';
      } elseif ($payingStatusNumber == 1) {
      $payingStatus = 'Paying';
      }
      preg_match_all($patternLastPayOut, $details, $payout);
      if (isset($payout[1][0])) {
      $payoutdate = $payout[1][0];
      } else {
      $payoutdate = ' Not Set';
      };
      $stmt2 = $conn->prepare('SELECT * FROM programs where :monitorNamePID=:id');
      $stmt2->bindParam('monitorNamePID', $monitorNamePID);
      $stmt2->bindParam('id', $id);
      $stmt2->execute();
      $numofupdates = $stmt2->rowCount();
      if ($numofupdates >= 1) {
      $stmt3 = $conn->prepare("UPDATE programs SET :monitorNamePayStatus=:payingstatus , :monitorNameLastPayout=:goldhyiplastpayout WHERE :monitorNamePID=:goldhyippid ");
      $stmt3->bindParam('monitorNamePayStatus', $monitorNamePayStatus);
      $stmt3->bindParam('monitorNameLastPayout', $monitorNameLastPayout);
      $stmt3->bindParam('monitorNamePID', $monitorNamePID);
      $stmt3->bindParam('payingstatus', $payingStatus);
      $stmt3->bindParam('goldhyiplastpayout', $payoutdate);
      $stmt3->bindParam('goldhyippid', $id);
      $stmt3->execute();
      echo 'P ID Updated : ' . $id . '<br>';
      } else {
      $siteAddress = get_redirect_final_host_url($gositesurl);
      echo $siteAddress;
      $stmt3 = $conn->prepare('INSERT INTO programs (:monitorNamePID, :monitorNamePayStatus, programName,:monitorNameLastPayout) VALUES (:goldhyippid, :payingstatus, :progname, :goldhyiplastpayout)');
      $stmt3->bindParam('monitorNamePID', $monitorNamePID);
      $stmt3->bindParam('monitorNamePayStatus', $monitorNamePayStatus);
      $stmt3->bindParam('monitorNameLastPayout', $monitorNameLastPayout);
      $stmt3->bindParam('goldhyippid', $id);
      $stmt3->bindParam('payingstatus', $payingStatus);
      $stmt3->bindParam('progname', $siteAddress);
      $stmt3->bindParam('goldhyiplastpayout', $payoutdate);
      $stmt3->execute();
      echo 'P ID inserted : ' . $id . '<br>';
      }
      }
      echo "Fetching $siteURL Done <br>";
      }


      I took details from sql, and add string to them, use in PDO query. When I write column name its ok but when I use Variable I got error.










      share|improve this question















      I want to get data from DB and add string to theme and put them in another query . When I run code I've got this error :




      Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or
      access violation: 1064 You have an error in your SQL syntax; check the
      manual that corresponds to your MySQL server version for the right
      syntax to use near ''goldhyipPID', 'goldhyipPayStatus',
      programName,'goldhyipLastPayout') VALUES ('1' at line 1 in
      C:wamp64wwwallmonitorstest.php on line 69




      <?php
      set_time_limit(3600);
      require_once 'fetchdetails/func.php';
      require_once 'config.php';
      $stmt = $conn->prepare("SELECT * FROM `monitors`");
      $stmt->execute();
      $result = $stmt->fetchAll(PDO::FETCH_ASSOC);
      foreach ($result as $monitor) {
      $monitorName = $monitor['monitorName'];
      $monitorNamePID = $monitorName . 'PID';
      $monitorNameLastPayout = $monitorName . 'LastPayout';
      $monitorNamePayStatus = $monitorName . 'PayStatus';
      $siteURL = $monitor['monitorurl'];
      $pattern1GetPID = $monitor['monitorPatternGetPID'];
      $patternLastPayOut = $monitor['monitorPatternLastPayout'];
      $patternPStatus = $monitor['monitorPatternPayStatus'];
      $patterndetailsurl = $monitor['monitorDetailsLink'];
      $patterngotositesurl = $monitor['monitorPatternGoSite'];
      $content = getPageContent($siteURL);
      preg_match_all($pattern1GetPID, $content, $matches, PREG_SET_ORDER, 0);
      foreach ($matches as $pid) {
      $id = $pid[1];
      $detailsurl = $patterndetailsurl . $id;
      $gositesurl = $patterngotositesurl . $id;
      $details = getPageContent($detailsurl);
      preg_match_all($patternPStatus, $details, $status);
      $payingStatusNumber = $status[1][0];
      if ($payingStatusNumber == 4) {
      $payingStatus = 'Not Paying';
      } elseif ($payingStatusNumber == 3) {
      $payingStatus = 'Problem';
      } elseif ($payingStatusNumber == 2) {
      $payingStatus = 'Waiting';
      } elseif ($payingStatusNumber == 1) {
      $payingStatus = 'Paying';
      }
      preg_match_all($patternLastPayOut, $details, $payout);
      if (isset($payout[1][0])) {
      $payoutdate = $payout[1][0];
      } else {
      $payoutdate = ' Not Set';
      };
      $stmt2 = $conn->prepare('SELECT * FROM programs where :monitorNamePID=:id');
      $stmt2->bindParam('monitorNamePID', $monitorNamePID);
      $stmt2->bindParam('id', $id);
      $stmt2->execute();
      $numofupdates = $stmt2->rowCount();
      if ($numofupdates >= 1) {
      $stmt3 = $conn->prepare("UPDATE programs SET :monitorNamePayStatus=:payingstatus , :monitorNameLastPayout=:goldhyiplastpayout WHERE :monitorNamePID=:goldhyippid ");
      $stmt3->bindParam('monitorNamePayStatus', $monitorNamePayStatus);
      $stmt3->bindParam('monitorNameLastPayout', $monitorNameLastPayout);
      $stmt3->bindParam('monitorNamePID', $monitorNamePID);
      $stmt3->bindParam('payingstatus', $payingStatus);
      $stmt3->bindParam('goldhyiplastpayout', $payoutdate);
      $stmt3->bindParam('goldhyippid', $id);
      $stmt3->execute();
      echo 'P ID Updated : ' . $id . '<br>';
      } else {
      $siteAddress = get_redirect_final_host_url($gositesurl);
      echo $siteAddress;
      $stmt3 = $conn->prepare('INSERT INTO programs (:monitorNamePID, :monitorNamePayStatus, programName,:monitorNameLastPayout) VALUES (:goldhyippid, :payingstatus, :progname, :goldhyiplastpayout)');
      $stmt3->bindParam('monitorNamePID', $monitorNamePID);
      $stmt3->bindParam('monitorNamePayStatus', $monitorNamePayStatus);
      $stmt3->bindParam('monitorNameLastPayout', $monitorNameLastPayout);
      $stmt3->bindParam('goldhyippid', $id);
      $stmt3->bindParam('payingstatus', $payingStatus);
      $stmt3->bindParam('progname', $siteAddress);
      $stmt3->bindParam('goldhyiplastpayout', $payoutdate);
      $stmt3->execute();
      echo 'P ID inserted : ' . $id . '<br>';
      }
      }
      echo "Fetching $siteURL Done <br>";
      }


      I took details from sql, and add string to them, use in PDO query. When I write column name its ok but when I use Variable I got error.







      php mysql pdo






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 11 at 5:44









      Tân Nguyễn

      1




      1










      asked Nov 11 at 5:39









      misagh ahmadi

      83




      83
























          2 Answers
          2






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          oldest

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          up vote
          0
          down vote













          Column names shouldn't be prepended with a colon. Remove colon : from your column names in both UPDATE and INSERT queries. Binding a table (or column) name doesn't work.



          $stmt3 = $conn->prepare("UPDATE programs SET monitorNamePayStatus=:payingstatus , monitorNameLastPayout=:goldhyiplastpayout WHERE monitorNamePID=:goldhyippid  ");

          $stmt3 = $conn->prepare('INSERT INTO programs (monitorNamePID, monitorNamePayStatus, programName, monitorNameLastPayout) VALUES (:goldhyippid, :payingstatus, :progname, :goldhyiplastpayout)');





          share|improve this answer






























            up vote
            0
            down vote



            accepted










            Binding a table (or column) name doesn't work






            share|improve this answer





















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              2 Answers
              2






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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

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              up vote
              0
              down vote













              Column names shouldn't be prepended with a colon. Remove colon : from your column names in both UPDATE and INSERT queries. Binding a table (or column) name doesn't work.



              $stmt3 = $conn->prepare("UPDATE programs SET monitorNamePayStatus=:payingstatus , monitorNameLastPayout=:goldhyiplastpayout WHERE monitorNamePID=:goldhyippid  ");

              $stmt3 = $conn->prepare('INSERT INTO programs (monitorNamePID, monitorNamePayStatus, programName, monitorNameLastPayout) VALUES (:goldhyippid, :payingstatus, :progname, :goldhyiplastpayout)');





              share|improve this answer



























                up vote
                0
                down vote













                Column names shouldn't be prepended with a colon. Remove colon : from your column names in both UPDATE and INSERT queries. Binding a table (or column) name doesn't work.



                $stmt3 = $conn->prepare("UPDATE programs SET monitorNamePayStatus=:payingstatus , monitorNameLastPayout=:goldhyiplastpayout WHERE monitorNamePID=:goldhyippid  ");

                $stmt3 = $conn->prepare('INSERT INTO programs (monitorNamePID, monitorNamePayStatus, programName, monitorNameLastPayout) VALUES (:goldhyippid, :payingstatus, :progname, :goldhyiplastpayout)');





                share|improve this answer

























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Column names shouldn't be prepended with a colon. Remove colon : from your column names in both UPDATE and INSERT queries. Binding a table (or column) name doesn't work.



                  $stmt3 = $conn->prepare("UPDATE programs SET monitorNamePayStatus=:payingstatus , monitorNameLastPayout=:goldhyiplastpayout WHERE monitorNamePID=:goldhyippid  ");

                  $stmt3 = $conn->prepare('INSERT INTO programs (monitorNamePID, monitorNamePayStatus, programName, monitorNameLastPayout) VALUES (:goldhyippid, :payingstatus, :progname, :goldhyiplastpayout)');





                  share|improve this answer














                  Column names shouldn't be prepended with a colon. Remove colon : from your column names in both UPDATE and INSERT queries. Binding a table (or column) name doesn't work.



                  $stmt3 = $conn->prepare("UPDATE programs SET monitorNamePayStatus=:payingstatus , monitorNameLastPayout=:goldhyiplastpayout WHERE monitorNamePID=:goldhyippid  ");

                  $stmt3 = $conn->prepare('INSERT INTO programs (monitorNamePID, monitorNamePayStatus, programName, monitorNameLastPayout) VALUES (:goldhyippid, :payingstatus, :progname, :goldhyiplastpayout)');






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 11 at 8:02

























                  answered Nov 11 at 7:55









                  Samir

                  5,1232626




                  5,1232626
























                      up vote
                      0
                      down vote



                      accepted










                      Binding a table (or column) name doesn't work






                      share|improve this answer

























                        up vote
                        0
                        down vote



                        accepted










                        Binding a table (or column) name doesn't work






                        share|improve this answer























                          up vote
                          0
                          down vote



                          accepted







                          up vote
                          0
                          down vote



                          accepted






                          Binding a table (or column) name doesn't work






                          share|improve this answer












                          Binding a table (or column) name doesn't work







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 20 at 17:20









                          misagh ahmadi

                          83




                          83






























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