1 
<?php $daerah_ejen1 = "$_GET[daerah_ejen]";
$kumpulan_ejen1 ="$_GET[kumpulan_ejen]";

echo $daerah_ejen1;
echo $kumpulan_ejen1;
echo $kumpulan_ejen;

$sql= "SELECT * FROM data_ejen WHERE daerah_ejen= '$daerah_ejen1' AND kumpulan_ejen='Ketua Kampung' ORDER BY nama_ejen";
$result = mysql_query($sql) or @error_die("Query failed : $sql " . mysql_error());
?>

my url

laporan_kk_detail.php?daerah_ejen=HULU+LANGAT&kumpulan_ejen=Ketua Kampung

for output daerah_ejen variable has display, but for kumpulan_ejen/kumpulan_ejen1 is not display.

i dont know where the problem

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1
  • You are echo $kumpulan_ejen; but you don't seem to have set $kumpulan_ejen anywhere. You've only set $kumpulan_ejen1 Commented Apr 2, 2014 at 7:23

5 Answers 5

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3

your quotes accessing $_GET variable is invalid. try this

<?php 
  $daerah_ejen1 = $_GET["daerah_ejen"];
  $kumpulan_ejen1 =$_GET["kumpulan_ejen"];

and you should read something about security, because you can pass malicous code to your script!

edit:// you can have a look on this thread https://stackoverflow.com/questions/19539692/sanitizing-user-input-php

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4 Comments

the first part of your answer is wrong. you can access the way the OP used.
but than, the 'daerah_ejen' for example will be treated as an constant, and since its not defined, its cast to a string. this is quite bad practise and raise notices within your application. so its just a lucky incident, that this actually works!
can you tell me what kind notice you was talking about ?
If your error_reporting and display_error php setting is switched on, you may see notice such as this: "Notice: Use of undefined constant daerah_ejen - assumed 'daerah_ejen'.
3

you are converting get values in string using double quotes so remove and try

$daerah_ejen1 = $_GET['daerah_ejen'];
$kumpulan_ejen1 =$_GET['kumpulan_ejen'];

also use mysql_real_escape_string() for prevent sql injection.

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0

  1. The quotes go around the parameter name. This is because $_GET[] is an associative array and its values are referenced using a string key

    $daerah_ejen1 = $_GET['daerah_ejen'];

    $kumpulan_ejen1 =$_GET['kumpulan_ejen'];

  2. Always sanitize your parameter values before using them in a query to protect yourself against SQL injection.

    $daerah_ejen1 = mysqli::real_escape_string($daerah_ejen1)

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0

You face 2 problem on your code :

1st is :

$daerah_ejen1 = "$_GET[daerah_ejen]";
$kumpulan_ejen1 ="$_GET[kumpulan_ejen]";

replace it by this :

 $daerah_ejen1 = $_REQUEST['daerah_ejen'];
$kumpulan_ejen1 =$_REQUEST['kumpulan_ejen'];

2nd is :

$sql= "SELECT * FROM data_ejen WHERE daerah_ejen= '$daerah_ejen1' AND kumpulan_ejen='Ketua Kampung' ORDER BY nama_ejen";

replace it by this :

$sql= "SELECT * FROM data_ejen WHERE daerah_ejen= '".$daerah_ejen1. "' AND kumpulan_ejen='Ketua Kampung' ORDER BY nama_ejen";
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7 Comments

can you tell me why you advice the OP to replace the $_GET to the way you have used ?
The second problem is not a problem. You can write "'$variable'" inside double quotes and get the desired result.
$_REQUEST represent $_GET and $_POST. I used $_REQUEST is easier your code which no need care about which $_GET or $_POST should use.
the question was asked before you change $_GET to $_REQUEST. And as said by W.K.S the second problem you mentioned was not a problem.Also changing to $_REQUEST will no way solve the problem
i recommend 2nd problem it more safety for your code.
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0

If you need to put the $_GET['name'] in double quotes, wrap it in {} brackets.

e.g. 

$kumpulan_ejen1 ="{$_GET['kumpulan_ejen']}";

Also, as dbh pointed out, you only have $kumpulan_ejen1, not kumpulan_ejen.

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