How to get values from MySQL database with PHP code using WHERE clause and variable set from url

Question

I want to get survey questions from MySQL database, where pollid=1 and echo them out.

I have following code:

<?php
include_once 'init/init.funcs.php';
$pollid=$_GET['pollid'];
$result = mysql_query('SELECT kysimus FROM kysimused where survey_id="1"');
$question = mysql_result($result, 0);
echo $pollid;
echo $question;

?>

It works and I see first question from database, where survey_id =1. But when I write:

<?php
include_once 'init/init.funcs.php';
$pollid=$_GET['pollid'];
$result = mysql_query('SELECT kysimus FROM kysimused where survey_id="$pollid"');
$question = mysql_result($result, 0);
echo $pollid;
echo $question;

?>

It says:

Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp2\htdocs\Praks\answering.php on line 5

In last example I inserted pollid from url and I am sure that variable $pollid got value '1'. Why the second code doesnt work? I have second question too, how could I create an array of all the questions, which have survey_id = $pollid.Thanks in advance.

Use double quotes if you want to pass variable into a statement. Double quotes for the statement and single quotes for the variable. — iamsleepy
– iamsleepy, Commented Apr 16, 2014 at 14:11
to elaborate on @iamsleepy 's answer, because you used single quotes it's trying to find the VALUE "$pollid" rather than the value contained within the variable. also, use parameter binding, mysqli or pdo, etc — Jeff Hawthorne
– Jeff Hawthorne, Commented Apr 16, 2014 at 14:14

Purefan · Accepted Answer · 2014-04-16 14:13:22Z

2

Single quote strings do not parse variables, change this: $result = mysql_query('SELECT kysimus FROM kysimused where survey_id="$pollid"');

for this: $result = mysql_query('SELECT kysimus FROM kysimused where survey_id="' . $pollid . '"');

Also dont forget to sanitize your input, something like this would be better: $result = mysql_query('SELECT kysimus FROM kysimused where survey_id="' . (int)$pollid . '"');

at least to make sure its an int

answered Apr 16, 2014 at 14:13

Purefan

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Comments

Olli · Accepted Answer · 2014-04-16 14:13:14Z

1

rewrite your query line as follows:

$result = mysql_query("SELECT kysimus FROM kysimused where survey_id='$pollid'");

This is due to the handling of single and double quotes in php.... The way you wrote it is the reason the variable does not get substituted.

Please see also: http://www.php.net/manual/en/language.types.string.php

more detailed: http://www.php.net/manual/en/language.types.string.php#language.types.string.parsing

answered Apr 16, 2014 at 14:13

Olli

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Comments

Adi Inbar · Accepted Answer · 2014-04-16 14:39:54Z

0

Try replacing this

$result = mysql_query('SELECT kysimus FROM kysimused where survey_id="$pollid"');

with

$result = mysql_query("SELECT kysimus FROM kysimused where survey_id='$pollid' ");

edited Apr 16, 2014 at 14:39

Adi Inbar

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answered Apr 16, 2014 at 14:14

echo_Me

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