can anyone help me !?

shahan

hi cools im new in php , i have 2 tables each table has diffrent data , i want make search for tables according to id or any other line virables , and for each table i want its result showing only , according to the selection in html ,
i have make below but when select second table i see some error \

this is html code

<html>

<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>Please dont live any blank space</title>
</head>

Please dont live any blank space while searching
Search Reservation Number 

<input type="text" name="search" size=25 maxlength=25 >

<select name="table" size="1">
<option selected="selected"></option>
<option value="fani">تهكنیكاران</option>
<option value="harafi">خزمهتگوزار</option>

   </select>
 <input type="Submit" name="Submit" value="Submit">
 </p>

</form>
</body>

</html>

and this is php code

<?php
$table=$POST['table'];
$search=$POST['search'];

if ($search == "" or $table=="")
{
echo "تۆ لهبیرت چووه ژمارهى چهكهكهت داخل بكهیت یاخود پۆلهكهت دیارى نهكردوه*";

echo"<a href='searchform.html'> كلیك لێره بكه بۆ گهرانهوه** </a>";
exit;
}

mysql_connect ("localhost", "root","root") or die (mysql_error());
mysql_select_db ("hawlerti");
mysql_query("SET NAMES 'utf8'");

{
if ($table=="fani")

$sql = mysql_query("select * from fani where chak like '$search'");

}
while ($row=mysql_fetch_array($sql))

{
echo 'ریزبهند: '.$row['id'];
echo ' ناوى سیانى: '.$row['name'];
echo ' ناونیشان: '.$row['nneshan'];
echo ' بروانامه: '.$row['brwanama'];
echo ' ژمارهى چهك: '.$row['chak'];
echo ' موژچهى بنهرهتى: '.$row['mbnarati'];
echo ' دهرمالهى خۆ تهرخانكردنى زانكۆیى %: '.$row['rzankoei'];
echo ' دةرمالةى خؤ تةرخانكردنى زانكؤيى: '.$row['zankoei'];
echo ' دهرمالهى برِوانامه* %: '.$row['rdbrwanama'];
echo ' ';

}
{
if ($table=="harafi")

$sql = mysql_query("select * from harafi where chak like '$search'");

}
while ($row=mysql_fetch_array($sql))

{
echo 'ریزبهند: '.$row['id'];
echo ' ناوى سیانى: '.$row['name'];
echo ' nawneshan: '.$row['nawneshan'];
echo ' بروانامه: '.$row['brwanama'];
echo ' ژمارهى چهك: '.$row['chak'];
echo ' موژچهى بنهرهتى: '.$row['mbnarati'];
echo ' دهرمالهى خۆ تهرخانكردنى زانكۆیى %: '.$row['rdzankoei'];
echo ' دةرمالةى خؤ تةرخانكردنى زانكؤيى: '.$row['zankoei'];
echo ' دهرمالهى برِوانامه* %: '.$row['rdbrwanama'];
echo ' ';
}

//This counts the number or results - and if there wasn't any it gives them a little message explaining that
$anymatches=mysql_num_rows($sql);
if ($anymatches == 0)

echo "ببوره ! ژمارهى چهكهكهت ههله داخل كردوه یاخود تۆ له پۆلى دیارى كراو نیت . تكایه ههڵهكان چاك بكهرهوهو دوباره ههوڵبده .";
echo"<a href='searchform.html'> كلیك لێره بكه بۆ گهرانهوه* </a>";

{

}

in second table when i make search in the result i see below error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\hawlertisearch\searchresult.php on line 29ریزبهند: 36
ناوى سیانى: باسم محمد علي
nawneshan: رئيس حرفين اقدم
بروانامه: ابتدائي
ژمارهى چهك: 2657
موژچهى بنهرهتى: 639000
دهرمالهى خۆ تهرخانكردنى زانكۆیى %: 1
دةرمالةى خؤ تةرخانكردنى زانكؤيى: 639000
دهرمالهى برِوانامه* %: 0.15

dalecosp

Hello, shahan, and welcome to PHPBuilder. Please use [noparse]

tags around your PHP code snippets and

tags around HTML and other code types[/noparse]; this allows better formatting and syntax highlighting, making it more convenient to help others with their questions.

The error (in red) is caused by the code below, I believe:

$sql = mysql_query("select * from harafi where chak like '$search'");

}
while ($row=mysql_fetch_array($sql))

You should check whether or not $sql is Boolean TRUE before attempting to run [man]mysql_fetch_array[/man]:

$sql = mysql_query("select * from harafi where chak like '$search'");

if ($sql) {
   while ($row=mysql_fetch_array($sql))

shahan

Rewrite entire code ,

Thanks a million Dear dalecosp really your helping so appreciate

Can you please copy my code and correct errors and past to me again until I use it directly bcz I triad with several changes without lucky ..

Thanks again ,

shahan

Dear Dalecosp ,
the problem is i use two times [ mysql_fetch_array ] in same file thats why when second table start to echo data the first $row=mysql_fetch_array show as error in line 29
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\hawlertisearch\searchresult.php on line 29