No Output in my php code

Question

Please help me.. I don't know why there is no output in my PHP code.. Below is my index.php code

<html>
<body>

<form action="search.php" method="POST">
<input type="text" name"number" placeholder="Search...."><br/>
<input type = "submit" value="Search">

</form>

</body>
</html>

and this is my search.php.

<html>
<body>
<?php

mysql_connect("localhost","root",""); 
mysql_select_db("login");

$initial = mysql_real_escape_string($_POST['initial']);

$find_videos = mysql_query("SELECT * FROM `sigup` WHERE `initial` LIKE '%initial%'");

while($row = mysql_fetch_assoc($find_videos))
{
$name = $row['name'];
echo "$name<br/ >";
}
?>
</body>
</html>

Are you sure your column name is `sigup` ? I doubt it is `signup`. — Shankar Narayana Damodaran, Nov 30 '13 at 12:33
On what basis do you assume that *search.php* should produce any output? — Oswald, Nov 30 '13 at 12:33
`echo mysql_error()` and `echo mysql_num_rows()` after the query. — Michael, Nov 30 '13 at 12:35
Alert!!! mysql_* functions are depreciated http://stackoverflow.com/q/12859942/829533 — zzlalani, Nov 30 '13 at 12:37

score 0 · Answer 1 · answered Nov 30 '13 at 12:33

0

 mysql_query("SELECT * FROM `sigup` WHERE `initial` LIKE '%$initial%'");

 <input type="text" name="initial" placeholder="Search....">

answered Nov 30 '13 at 12:33

Maresh

4,644
25
30

Guru · Answer 2 · 2013-12-02T05:36:07.030

0

<html>
<body>

<form action="search.php" method="POST">
<input type="text" name="initial" placeholder="Search...."><br/>
<input type = "submit" value="Search">

</form>

</body>
</html>

search.php

<html>
<body>
<?php

mysql_connect("localhost","root",""); 
mysql_select_db("login");

$initial = mysql_real_escape_string($_POST['initial']);

$find_videos = mysql_query("SELECT * FROM `sigup` WHERE `initial` LIKE '%$initial%'");

while($row = mysql_fetch_assoc($find_videos))
{
$name = $row['name'];
echo "$name<br/ >";
}
?>
</body>
</html>

edited Dec 02 '13 at 05:36

answered Nov 30 '13 at 12:41

Guru

621
1
4
12

but this the another error "Notice: Undefined index: initial in C:\wamp\www\ralph\search.php on line 9" .. :(( – user3051787 Nov 30 '13 at 12:59
what is initial in search.php – Guru Nov 30 '13 at 13:00
i changed the name of the button
........ and $initial = mysql_real_escape_string($_POST['number']); i guess that ($_POST['number']) is the name of button.. – user3051787 Nov 30 '13 at 13:11
but it say's "Notice: Undefined index: number in C:\wamp\www\ralph\search.php on line 6" .. :(( – user3051787 Nov 30 '13 at 13:12
try to remove mysql_real_escape_string – Guru Nov 30 '13 at 13:15
please change'%$initial%' instead of '%initial%'.definatly show output. – Guru Dec 02 '13 at 05:35

score 0 · Answer 3 · answered Nov 30 '13 at 12:54

    <html>
    <body>

    <form action="search.php" method="POST">
    <input type="text" name="number" placeholder="Search...."><br/>
     <input type = "submit" value="Search">

    </form>

  </body>
</html>

search.php

    <html>
    <body>
    <?php
    mysql_connect("localhost","root",""); 
    mysql_select_db("login");
    $initial = mysql_real_escape_string($_POST['number']);
    $find_videos = mysql_query("SELECT * FROM sigup  WHERE initial LIKE '%initial%'");
    while($row = mysql_fetch_assoc($find_videos))
    {
    $name = $row['name'];
    echo "$name";
    }
    ?>
    </body>
   </html>

No Output in my php code

3 Answers3