Reputation:
I have to two tables in my database one is "test1" and the second is "test2". and the field for test 1 is "id","survey_no","lastname" and for the field of test i have "id","survey_no","address" the id for test1 and test2 are primary key and set to auto-increment..what i want here is that what ever i insert in "survey_no" in test1 should also be insert inserted in "survey_no in test2"..can somebody please help me with it..please..
I have here the code for insertion called sample.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form method="POST" class="signin" action="" name="add" id="form1">
<fieldset class="textbox">
<label class="province_id">
<span>province id</span>
<input id="province" name="survey_no" type="number" value="" autocomplete="on" placeholder="survey_no">
</label>
<label class="municipality">
<span>Municipality</span>
<input id="municipality" name="lastname" value="" type="text" autocomplete="on" placeholder="lastname">
</label>
<br />
<br />
<button id="submit" type="submit" name="submit">Save</button>
<button id="submit" type="reset" name="reset">Reset</button>
</fieldset>
<?php
$conn=mysql_connect('localhost','root','');
if(!$conn)
{
die('could not connect:' .mysql_error());
}
mysql_select_db("sample",$conn);
if(isset($_POST['submit']))
{
$survey_no=$_POST['survey_no'];
$lastname=$_POST['lastname'];
$result = mysql_query("INSERT INTO `test1`(survey_no,lastname)
VALUES ('$survey_no','$lastname')");
if($result)
{
echo "<script type=\"text/javascript\">".
"alert('Saved!');".
"</script>";
}
else{
echo "<script type=\"text/javascript\">".
"alert('Failed!');".
"</script>";
}
}
?>
</form>
</body>
</html>
Upvotes: 2
Views: 89
Reputation: 2762
so far i understand you want this
$result = mysql_query("INSERT INTO `test1`(survey_no,lastname)
VALUES ('$survey_no','$lastname')");
$result1 = mysql_query("INSERT INTO `test2`(survey_no)
VALUES ('$survey_no')");
Upvotes: 4