POST variable always stays empty
up vote
-1
down vote
favorite
This is the method in which i manage the data send with jquery.ajax. By default is send an empty string and on each change i watch the change on the input and resend it. Through console things are good but in php $this->searchField always has the value of an empty string
function checkInfo(){
if(isset($_POST['searchField'])){
$this->searchField = json_decode($_POST['searchField']);
if ($this->searchField != ""){
$this->query = "SELECT * FROM myguests WHERE firstName like '%".$searchField."%'";
}
else if ($this->searchField == "") {
$this->query = "SELECT * FROM myguests ORDER BY id";
}
$this->tableDisplay();
exit();
}
else{
return "error";
}
}
This is my jquery ajax function :
function searchGuests(users){
var data = {
"searchField": users
};
console.log(data);
$.ajax({
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
})
.done(function(tr) {
$("#tableBody").empty();
$("#tableBody").append(tr);
})
.fail(function(error){
console.log(error);
});
};
Any idea what I am missing?
javascript php jquery sql ajax
asked Nov 11 at 11:31
juxhin bleta
598
add a comment |
up vote
-1
down vote
favorite
This is the method in which i manage the data send with jquery.ajax. By default is send an empty string and on each change i watch the change on the input and resend it. Through console things are good but in php $this->searchField always has the value of an empty string
function checkInfo(){
if(isset($_POST['searchField'])){
$this->searchField = json_decode($_POST['searchField']);
if ($this->searchField != ""){
$this->query = "SELECT * FROM myguests WHERE firstName like '%".$searchField."%'";
}
else if ($this->searchField == "") {
$this->query = "SELECT * FROM myguests ORDER BY id";
}
$this->tableDisplay();
exit();
}
else{
return "error";
}
}
This is my jquery ajax function :
function searchGuests(users){
var data = {
"searchField": users
};
console.log(data);
$.ajax({
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
})
.done(function(tr) {
$("#tableBody").empty();
$("#tableBody").append(tr);
})
.fail(function(error){
console.log(error);
});
};
Any idea what I am missing?
javascript php jquery sql ajax
asked Nov 11 at 11:31
juxhin bleta
598
what does the users variable holding?
– Ali
Nov 11 at 11:50
1
You aren't sending any JSON so no need to use json_decode()
– charlietfl
Nov 11 at 11:54
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
This is the method in which i manage the data send with jquery.ajax. By default is send an empty string and on each change i watch the change on the input and resend it. Through console things are good but in php $this->searchField always has the value of an empty string
function checkInfo(){
if(isset($_POST['searchField'])){
$this->searchField = json_decode($_POST['searchField']);
if ($this->searchField != ""){
$this->query = "SELECT * FROM myguests WHERE firstName like '%".$searchField."%'";
}
else if ($this->searchField == "") {
$this->query = "SELECT * FROM myguests ORDER BY id";
}
$this->tableDisplay();
exit();
}
else{
return "error";
}
}
This is my jquery ajax function :
function searchGuests(users){
var data = {
"searchField": users
};
console.log(data);
$.ajax({
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
})
.done(function(tr) {
$("#tableBody").empty();
$("#tableBody").append(tr);
})
.fail(function(error){
console.log(error);
});
};
Any idea what I am missing?
javascript php jquery sql ajax
asked Nov 11 at 11:31
juxhin bleta
598
This is the method in which i manage the data send with jquery.ajax. By default is send an empty string and on each change i watch the change on the input and resend it. Through console things are good but in php $this->searchField always has the value of an empty string
function checkInfo(){
if(isset($_POST['searchField'])){
$this->searchField = json_decode($_POST['searchField']);
if ($this->searchField != ""){
$this->query = "SELECT * FROM myguests WHERE firstName like '%".$searchField."%'";
}
else if ($this->searchField == "") {
$this->query = "SELECT * FROM myguests ORDER BY id";
}
$this->tableDisplay();
exit();
}
else{
return "error";
}
}
This is my jquery ajax function :
function searchGuests(users){
var data = {
"searchField": users
};
console.log(data);
$.ajax({
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
})
.done(function(tr) {
$("#tableBody").empty();
$("#tableBody").append(tr);
})
.fail(function(error){
console.log(error);
});
};
Any idea what I am missing?
javascript php jquery sql ajax
javascript php jquery sql ajax
asked Nov 11 at 11:31
juxhin bleta
598
asked Nov 11 at 11:31
juxhin bleta
598
asked Nov 11 at 11:31
juxhin bleta
598
asked Nov 11 at 11:31
juxhin bleta
598
asked Nov 11 at 11:31
juxhin bleta
598
598
what does the users variable holding?
– Ali
Nov 11 at 11:50
1
You aren't sending any JSON so no need to use json_decode()
– charlietfl
Nov 11 at 11:54
add a comment |
what does the users variable holding?
– Ali
Nov 11 at 11:50
1
You aren't sending any JSON so no need to use json_decode()
– charlietfl
Nov 11 at 11:54
what does the users variable holding?
– Ali
Nov 11 at 11:50
what does the users variable holding?
– Ali
Nov 11 at 11:50
1
1
You aren't sending any JSON so no need to use json_decode()
– charlietfl
Nov 11 at 11:54
You aren't sending any JSON so no need to use json_decode()
– charlietfl
Nov 11 at 11:54
add a comment |
1 Answer
1
active
oldest
votes
up vote
-1
down vote
Try encoding the users to a string before sending it over.
function searchGuests(users){
var data = {
"searchField": JSON.stringify(users);
};
console.log(data);
$.ajax({
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
})
.done(function(tr) {
$("#tableBody").empty();
$("#tableBody").append(tr);
})
.fail(function(error){
console.log(error);
});
};
answered Nov 11 at 11:46
Edwin Dijas Chiwona
34117
No such method asJSON.encode()and the idea doesn't make sense either. jQuery uses$.param()to serialize objects by default
– charlietfl
Nov 11 at 11:52
I solved it indeed it didn't needed to be json_decode -ed
– juxhin bleta
Nov 11 at 11:52
confused it with php json_encode. it is JSON.stringify
– Edwin Dijas Chiwona
Nov 11 at 12:02
please share your answer. Thanks
– Edwin Dijas Chiwona
Nov 11 at 12:07
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
Try encoding the users to a string before sending it over.
function searchGuests(users){
var data = {
"searchField": JSON.stringify(users);
};
console.log(data);
$.ajax({
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
})
.done(function(tr) {
$("#tableBody").empty();
$("#tableBody").append(tr);
})
.fail(function(error){
console.log(error);
});
};
answered Nov 11 at 11:46
Edwin Dijas Chiwona
34117
No such method asJSON.encode()and the idea doesn't make sense either. jQuery uses$.param()to serialize objects by default
– charlietfl
Nov 11 at 11:52
I solved it indeed it didn't needed to be json_decode -ed
– juxhin bleta
Nov 11 at 11:52
confused it with php json_encode. it is JSON.stringify
– Edwin Dijas Chiwona
Nov 11 at 12:02
please share your answer. Thanks
– Edwin Dijas Chiwona
Nov 11 at 12:07
add a comment |
up vote
-1
down vote
Try encoding the users to a string before sending it over.
function searchGuests(users){
var data = {
"searchField": JSON.stringify(users);
};
console.log(data);
$.ajax({
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
})
.done(function(tr) {
$("#tableBody").empty();
$("#tableBody").append(tr);
})
.fail(function(error){
console.log(error);
});
};
answered Nov 11 at 11:46
Edwin Dijas Chiwona
34117
No such method asJSON.encode()and the idea doesn't make sense either. jQuery uses$.param()to serialize objects by default
– charlietfl
Nov 11 at 11:52
I solved it indeed it didn't needed to be json_decode -ed
– juxhin bleta
Nov 11 at 11:52
confused it with php json_encode. it is JSON.stringify
– Edwin Dijas Chiwona
Nov 11 at 12:02
please share your answer. Thanks
– Edwin Dijas Chiwona
Nov 11 at 12:07
add a comment |
up vote
-1
down vote
up vote
-1
down vote
Try encoding the users to a string before sending it over.
function searchGuests(users){
var data = {
"searchField": JSON.stringify(users);
};
console.log(data);
$.ajax({
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
})
.done(function(tr) {
$("#tableBody").empty();
$("#tableBody").append(tr);
})
.fail(function(error){
console.log(error);
});
};
answered Nov 11 at 11:46
Edwin Dijas Chiwona
34117
Try encoding the users to a string before sending it over.
function searchGuests(users){
var data = {
"searchField": JSON.stringify(users);
};
console.log(data);
$.ajax({
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
})
.done(function(tr) {
$("#tableBody").empty();
$("#tableBody").append(tr);
})
.fail(function(error){
console.log(error);
});
};
answered Nov 11 at 11:46
Edwin Dijas Chiwona
34117
edited Nov 11 at 12:06
answered Nov 11 at 11:46
Edwin Dijas Chiwona
34117
answered Nov 11 at 11:46
Edwin Dijas Chiwona
34117
answered Nov 11 at 11:46
Edwin Dijas Chiwona
34117
34117
No such method asJSON.encode()and the idea doesn't make sense either. jQuery uses$.param()to serialize objects by default
– charlietfl
Nov 11 at 11:52
I solved it indeed it didn't needed to be json_decode -ed
– juxhin bleta
Nov 11 at 11:52
confused it with php json_encode. it is JSON.stringify
– Edwin Dijas Chiwona
Nov 11 at 12:02
please share your answer. Thanks
– Edwin Dijas Chiwona
Nov 11 at 12:07
add a comment |
No such method asJSON.encode()and the idea doesn't make sense either. jQuery uses$.param()to serialize objects by default
– charlietfl
Nov 11 at 11:52
I solved it indeed it didn't needed to be json_decode -ed
– juxhin bleta
Nov 11 at 11:52
confused it with php json_encode. it is JSON.stringify
– Edwin Dijas Chiwona
Nov 11 at 12:02
please share your answer. Thanks
– Edwin Dijas Chiwona
Nov 11 at 12:07
No such method as
JSON.encode() and the idea doesn't make sense either. jQuery uses $.param() to serialize objects by default– charlietfl
Nov 11 at 11:52
No such method as
JSON.encode() and the idea doesn't make sense either. jQuery uses $.param() to serialize objects by default– charlietfl
Nov 11 at 11:52
I solved it indeed it didn't needed to be json_decode -ed
– juxhin bleta
Nov 11 at 11:52
I solved it indeed it didn't needed to be json_decode -ed
– juxhin bleta
Nov 11 at 11:52
confused it with php json_encode. it is JSON.stringify
– Edwin Dijas Chiwona
Nov 11 at 12:02
confused it with php json_encode. it is JSON.stringify
– Edwin Dijas Chiwona
Nov 11 at 12:02
please share your answer. Thanks
– Edwin Dijas Chiwona
Nov 11 at 12:07
please share your answer. Thanks
– Edwin Dijas Chiwona
Nov 11 at 12:07
add a comment |
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what does the users variable holding?
– Ali
Nov 11 at 11:50
1
You aren't sending any JSON so no need to use json_decode()
– charlietfl
Nov 11 at 11:54