Hi , How can I show CaseID once user has fill up the form .
My database : CaseID , Name , Feedback
The form will require user to enter name and feedback . After submit button is enter , the CaseID should be show to user .
How can I do it ? I am using xampp as database.
How can I do it ? I am using xampp as database.
@erminasrcutp
How do you want it? PHP or jQuery or javascript?
anything as long as it works ..
PHP will be given priority
PHP will be given priority
@erminasrcutp
Do you know or familiar with PHP at ?
yes :) well i manage to insert data to database myqsl however i dont manage to show user the caseID which is auto increment
yes :) well i manage to insert data to database myqsl however i dont manage to show user the caseID which is auto increment
Post your INSERT query, then we can go from there
<!DOCTYPE>
<html>
<head>
<title>feedback</title>
</head>
<div id="header">
<img src='banner.png' align='middle'>
</div>
<div id="navigation">
<div class="mainMenu">
<ul><li>
<a echo ; href='main.php' >Main</a>
</li><li>
<a echo class='selected'; href='feedback.php'>
New feedback</a>
</div>
</div>
<div id="content">
<form name="borang" method="post"
onsubmit="return checkField()" enctype="multipart/form-data">
<table align="center" id="table" >
<th colspan="2" class="th">feedback form</th>
<tr>
<td class="td" colspan="2">
<br/>
have to be fill
<span class="span">*</span>
</td>
</tr>
<tr>
<td class="td">
<br/>
name
</td>
<td>
<br/>
<input type="text" name="name" value="" />
</td>
</tr>
<tr>
<td class="td">
feedback
</td>
<td>
<input type="text" name="feedback" value="" />
<span class="span">*</span>
</td>
</tr>
<tr>
<td class="td">
</td>
<td class="td" >
<br/>
<button type="submit" value="submit"name="submit"/>
check
</button>
<button type="reset" value="reset"/>
reset
</button>
<br/>
</td>
</tr>
</table>
<br><br>
<?php
if (isset($_POST ['submit'])){
$con = mysql_connect("localhost","root","");
if (!$con)
{
die("Can not connect ".mysql_error());
}
mysql_select_db("database",$con);
$sql = "INSERT INTO aduan(name,feedback)
VALUES('$_POST[name]','$_POST[feedback]')";
mysql_query($sql,$con);
mysql_close($con);
}
?>
</form>
</div>
<div id="footer">
Bye bye
</div>
</body>
</html>
you probably need to use this function: http://pk1.php.net/mysql_insert_id
have you looked at it before?
Let me know if you need further help about where to call this function and to display the id of newly inserted row that it will return to the user in subsequent response.
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