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update mysql form select value

Thread began 4/21/2018 11:46 am by anonymous | Last modified 4/23/2018 12:53 pm by Ray Borduin | 398 views | 1 replies |

anonymous

update mysql form select value

hi
i want update data mysql to select value
just no update?
please help me?


if(isset($_POST['Submit']))
{
$sql=("SELECT * FROM dars
WHERE

major like '%$major%'
and sal like '%$sal%'
and type2 like '%$type2%'
and type like '%$type%'
and title like '$title%'
and vahed like '%$vahed%'
and nimsal like '%$nimsal%'");
$result=mysql_query($sql,$con);
$n=1;
<?php
if (mysql_num_rows($result) > 0)
{
?>

<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">

<?php
while($row=mysql_fetch_assoc($result))
{
echo $n++ . " ----- " //. " darsid: "?>
<input type="hidden" name="darsid[]" value="<?php $row['darsid']; ?>"/>

sal : <select name="sal[]">
<option value="اول دبیرستان"<?php if($row['sal']=='اول دبیرستان') echo 'selected="selected"'; ?>>اول دبیرستان </option>
<option value="دوم دبیرستان"<?php if($row['sal']=='دوم دبیرستان') echo 'selected="selected"'; ?>>دوم دبیرستان </option>
<option value="سوم دبیرستان"<?php if($row['sal']=='سوم دبیرستان') echo 'selected="selected"'; ?>>سوم دبیرستان </option>
</select>
major : <select name="major[]">
<option value="کامپیوتر"<?php if($row['major']=='کامپیوتر') echo 'selected="selected"'; ?>>کامپیوتر </option>
<option value="حسابداری"<?php if($row['major']=="حسابداری")echo 'selected="selected"'; ?>>حسابداری </option>
<option value="برق"<?php if($row['major']=='برق') echo 'selected="selected"'; ?>>برق </option>
</select>
type2: <select name="type2[]">
<option value="عمومی"<?php if($row['type2']=='عمومی') echo 'selected="selected"'; ?>>عمومی </option>
<option value="تخصصی"<?php if($row['type2']=='تخصصی') echo 'selected="selected"'; ?>> تخصصی</option>
</select>
type: <select name="type[]">
<option value="تئوری"<?php if($row['type']=='تئوری') echo 'selected="selected"'; ?>>تئوری </option>
<option value="عملی"<?php if($row['type']=='عملی') echo 'selected="selected"'; ?>> عملی</option>
</select>
title: <input type="text" name="title[]" value="<?php echo $row['title']; ?>"/>
vahed: <select name="vahed[]">
<option value="1"<?php if($row['vahed']=='1') echo 'selected="selected"'; ?>>1</option>
<option value="2"<?php if($row['vahed']=='2') echo 'selected="selected"'; ?>>2 </option>
<option value="3"<?php if($row['vahed']=='3') echo 'selected="selected"'; ?>>3 </option>
<option value="4"<?php if($row['vahed']=='4') echo 'selected="selected"'; ?>>4 </option>
</select>
nimsal <select name="nimsal[]">
<option value="1"<?php if($row['nimsal']=='1') echo 'selected="selected"'; ?>>نیمسال اول</option>
<option value="2"<?php if($row['nimsal']=='2') echo 'selected="selected"'; ?>>نیمسال دوم </option>
<option value="3"<?php if($row['nimsal']=='3') echo 'selected="selected"'; ?>>کامل </option>
</select>

<?php

echo "<br>";



}
}


}?>
<input type="submit" name="edit" value="edit">
</form>
<?php

if(isset($_POST['edit']))
{

$size = count($_POST['darsid']);
$i = 0;
while ($i < $size) {
$id1=$_POST['darsid'][$i];
$sal1=$_POST['sal'][$i];
$major1=$_POST['major'][$i];
$type21=$_POST['type2'][$i];
$type1=$_POST['type'][$i];
$title1=$_POST['title'][$i];
$vahed1=$_POST['vahed'][$i];
$nimsal1=$_POST['nimsal'][$i];

$sql2="Update dars set
sal='$sal1',major='$major1',type2='$type21',type='$type1',title='$title1',vahed='$vahed1',
nimsal='$nimsal1'
Where darsid='$id1'";

mysql_query($sql2) or die ("Error in query: $query");


++$i;
}
}
?>

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Ray BorduinWebAssist

The webassist DataBridge extension can help you create recordsets and update code instead of using hand coding. I can see your hand code has SQL Injection vulnerabilities and it is difficult for me to debug remotely. If you wanted to use a premier support ticket I could help you debug what is wrong with your hand written code.

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