value of echo not appear.

Question

<?php
include("db/dbconnect2.php");

$noMyKid1 = $_GET['noMyKid'];

echo $noMyKid1;

$sql= "SELECT namaAnak,noSijilLahir
FROM mohon a 
INNER JOIN tblstatus_tak_lengkap b on a.noMyKid=b.nomykid 
WHERE a.noMyKid='noMyKid1'";
$result = mysql_query($sql) or @error_die("Query failed : $sql " . mysql_error());

echo $noMyKid;
echo $namaAnak;
echo $noSijilLahir;
?>

I run this coding, but the output just appear for

echo $noMyKid1;,echo $noMyKid;

Why echo $namaAnak;and echo $noSijilLahir; not appear?

In the SQL query you probably meant to say `WHERE a.noMyKid='$noMyKid1'` (you should add the $ if you are using the variable). Also, you run the query, which goes into `$result`, but you don't do anything with the result. It does not auto-populate variables from your query. You need to fetch the values first. — Always Learning, Apr 14 '14 at 04:12
[Why shouldn't I use mysql_* functions in PHP?](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php) — Phil, Apr 14 '14 at 04:13

potashin · Accepted Answer · 2014-04-14T04:18:13.863

2

Cause they are not defined. Also, you've forgotten $ sign(I guess, if there is a.noMyKid='noMyKid1' then sorry):

$sql= "SELECT namaAnak,noSijilLahir
FROM mohon a 
INNER JOIN tblstatus_tak_lengkap b on a.noMyKid=b.nomykid 
WHERE a.noMyKid='$noMyKid1'";

And to fetch you result set:

while($row = mysql_fetch_assoc($result)){
   echo $row['namaAnak'];
   echo $row['noSijilLahir'];
}

edited Apr 14 '14 at 04:18

answered Apr 14 '14 at 04:11

potashin

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yep, he/she did not fetch the data from the query result. – DoubleM Apr 14 '14 at 04:13
thank you so much. I realize the fetch data before i read the comment. but i forgot to put $ sign.LOL~. thank you for helping.. – user3487681 Apr 14 '14 at 04:17

score 0 · Answer 2 · answered Apr 14 '14 at 04:16

Because you are echoing empty variable without initialized any value on it:

This is full code may be working for you:

<?php
include("db/dbconnect2.php");

$noMyKid1 = $_GET['noMyKid'];

echo $noMyKid1;

$sql= "SELECT namaAnak,noSijilLahir
FROM mohon a 
INNER JOIN tblstatus_tak_lengkap b on a.noMyKid=b.nomykid 
WHERE a.noMyKid='$noMyKid1'";
$result = mysql_query($sql) or @error_die("Query failed : $sql " . mysql_error());

while ($row = mysql_fetch_array($result))
{
   echo $row['namaAnak'];
   echo $row['noSijilLahir'];
}
?>

And you have forgotten $ sign on your query string

WHERE a.noMyKid='$noMyKid1'";

value of echo not appear.

2 Answers2