CODEWITHSUNDEEP
phpmysqlajax
Daniel Developer
Daniel Developer

Reputation: 9

Database query failedYou have an error in your SQL syntax;

here is the Error Database query failedYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' )' at line 4..

im so tired to fixed this error and i dont know what should i do..

<form id="packageLessonType" method="post">
<table id="listpackage" class="table table-condensed table-hover">
<thead>
<tbody id="responsePackage">
<tr id="pack_1">
<td>1 Month Unlimited</td>
<td id="td-" class="center">
<input type="checkbox" value="1" name="package[]">
</td>
</tr>
<tr id="pack_2">
<td>6 Lessons</td>
<td id="td-" class="center">
<input type="checkbox" value="2" name="package[]">
</td>
</tr>
<tr id="pack_3">
<td>6 Months Unlimited</td>
<td id="td-" class="center">
<input type="checkbox" value="3" name="package[]">
</td>
</tr>
<tr id="pack_4">
<td>12 Months Unlimited</td>
<td id="td-" class="center">
<input type="checkbox" value="4" name="package[]">
</td>
</tr>
<tr id="pack_5">
<td>Group of 4</td>
<td id="td-" class="center">
<input type="checkbox" value="5" name="package[]">
</td>
</tr>
<tr id="pack_6">
<td>Group of 8 </td>
<td id="td-" class="center">
<input type="checkbox" value="6" name="package[]">
</td>
</tr>
</tbody>
</table>
</form>

Ajax Script

$('.packAddLt_btn').click(function(e){
        var clickedIdadded = this.id.split('_');
        var addedId = clickedIdadded[1];
        var addedPackageLesson = "&lessonPackageAdded="+addedId;
        var addLesPackCheck = $('#packageLessonType').serialize();
        var submitData = addLesPackCheck+addedPackageLesson;
        //alert(submitData);

        $.ajax({
            type:"POST",
            url:"<?php echo get_stylesheet_directory_uri(); ?>/includes/response_lessons_packages.php",
            dataType:"text",
            data:submitData,
            success: function(response){

                alert(response);
                },
            error: function(xhr, ajaxOtions, thrownError){
                    alert(thrownError);
                }
            });

        });

PHP CODE

if($_POST['package']){

            $checkBox= $_POST['package'];
            $idToBeAdded = $_POST['lessonPackageAdded'];

            foreach($checkBox as $value){
                $queryAdd = "INSERT INTO packagebylessontype (packageid)VALUE({$value}) WHERE lessontypeid = {$idToBeAdded}";
                $addedLessonByPackage = mysql_query($queryAdd, $connection);
                }
                confirm_query($addedLessonByPackage);

            if($addedLessonByPackage){
                echo "You've successfuly deleted your Package..";
                mysql_close($connection);
                }else{
                header('HTTP/1.1 500 Looks like mysql error, could not insert record!');
                exit();

                }   

    }

Error:

Database query failedYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' )' at line 4

Upvotes: 0

Views: 81

Answers (4)

manuelbcd
manuelbcd

Reputation: 4577

You cannot use INSERT statement with WHERE clause in a simple query. An INSERT will create a new row with his own field lessontypeid. Maybe you are trying to perform an update.

If you want to make an insert, the correct way would be:

INSERT INTO packagebylessontype (packageid) VALUES ('A')

If you want to perform an update, then:

UPDATE packagebylessontype SET packageid = 'A' WHERE lessontypeid ='1'

Please note that I'm using WHERE clause only in UPDATE operation, as a reference to locate register to be updated.

Upvotes: 2

Semicolon
Semicolon

Reputation: 1945

try with VALUES instead of VALUE $queryAdd = "INSERT INTO packagebylessontype (packageid)VALUES({$value}) WHERE lessontypeid = {$idToBeAdded}";

Upvotes: 0

Ali Shaan
Ali Shaan

Reputation: 345

Well give a space between (packageid) and VALUES, I think this might work and ofcourse it should be VALUES not VALUE

Upvotes: 0

Martin Kovachev
Martin Kovachev

Reputation: 442

You have a syntax error - instead of VALUE type VALUES.

Also you are not properly escaping your sql params thus leading yourself open to sql injections.

Upvotes: 2

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