Using WHERE function with $_POST

Question

I am currently importing data using the $_POST function. I want to use the imported value as the selected column. Currenlty my code looks like this:

$value = $_POST["name"]

$sql = "SELECT question_id, question FROM Question WHERE question_id= $value";

I am currently getting an error when I call my php script using JQuery. Is this section of the code the potential source?

This is the JQuery code I'm using.

<script type="text/javascript">
    $(document).ready(function(){
        $('button').click(function(){
            $('#show').load('/php/getQuestion.php'), { name: $("#txtname").val()}
        });

    });
</script>

Txtname is <strong>Name:</strong><input type = "text" id = "txtname">

There error is

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I think problem is quote. Use '(single) quote to your variable. $sql = "SELECT question_id, question FROM Question WHERE question_id= '".$value."'"; ...... — Jaydeep Mor
– Jaydeep Mor, Commented Jul 3, 2017 at 4:11
check the field type of question_id column, i think it should be question_id= '$value' — Rahul
– Rahul, Commented Jul 3, 2017 at 4:11
What is the error message? You are open to sql injection, you should sanitize the posted data. — Sean
– Sean, Commented Jul 3, 2017 at 4:11

Rohan Kumar · Accepted Answer · 2017-07-03 04:33:26Z

1

Your $.load() closing is not correct and it takes second parameter as arguments, but you are closing it before passing your params

$(document).ready(function(){
    $('button').click(function(){
        $('#show').load('/php/getQuestion.php',
          { name: $("#txtname").val()} );
               // ---------------------^ closing parenthesis of load()
    });
});

In PHP you have missed semicolon after $value = $_POST["name"]

answered Jul 3, 2017 at 4:33

Rohan Kumar

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Comments

Amaan Iqbal · Accepted Answer · 2017-07-03 04:16:28Z

1

Try

$sql = "SELECT question_id, question FROM Question WHERE question_id = ".$value;

Hope it helps

answered Jul 3, 2017 at 4:16

Amaan Iqbal

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Comments

Chintan7027 · Accepted Answer · 2017-07-03 04:30:39Z

0

First of all wrap your argument as string

$value = $_POST["name"];
$sql = "SELECT `question_id`, `question` FROM Question WHERE `question_id` like '%$value%'";

or

$value = (integer)$_POST["name"];
$sql = "SELECT `question_id`, `question` FROM Question WHERE `question_id` = {$value}";

edited Jul 3, 2017 at 4:30

answered Jul 3, 2017 at 4:12

Chintan7027

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8 Comments

Murad Hasan Over a year ago

Why wrap?? If the question _id is int then it will not necessary to wrap. Why use like?

Chintan7027 Over a year ago

Yes agree integere value need to wrap, but he don'e mention anything that's why using anything from $_POST is not safe

Chintan7027 Over a year ago

If sure that id is integer than cast the $_POST value to integer at either PHP level or mysql level

Murad Hasan Over a year ago

So you may ask the OP about your clearance, no need to post it as answer which is actually not the valid answer.

user02020202 Over a year ago

I wanted an integer, sorry for not specifying.

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