Hi, I recently created a mysql database using phpmyadmin. I then proceeded to make a form to insert data into the database, but the problem is that the form is only able to insert one record, and then if I try inserting another record, the new one is not seen in the database. There is no error messages seen in the form when entering a new record.
Heres the code for the form -
<html>
-
<head>
-
</head>
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<center>
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<form method="post" action="script.php">
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<input type="hidden" name="id" value="null">
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<table>
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<tr><td align="left">Name</td>
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<td><input type="text" name="name"></td>
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</tr>
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<tr><td align="left">Telephone</td>
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<td><input type="text" name="telephone" size="20"></td>
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</tr>
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<tr><td align="left">Birthday</td>
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<td><input type="text" name="birthday" size="20"></td>
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</tr>
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<tr><td colspan="2">
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<p align="center">
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<input type="submit" value="Enter record">
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</td>
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</tr>
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</table>
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</form>
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</center>
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</html>
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And Heres the code for the php script -
<?
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$DBhost = "localhost";
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$DBuser = "username";
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$DBpass = "password";
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$DBName = "database name";
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$table = "table name";
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mysql_connect($DBhost,$DBuser,$DBpass) or die("Unable to connect to database");
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@mysql_select_db("$DBName") or die("Unable to select database $DBName");
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$sqlquery = "INSERT INTO $table VALUES('$id','$name','$telephone','$birthday')";
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$results = mysql_query($sqlquery);
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mysql_close();
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print "<html><body><center>";
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print "<p>You have just entered this record<p>";
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print "Name : $name<br>";
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print "Telephone : $telephone<br>";
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print "Birthday :$birthday";
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print "</body></html>";
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?>
5  2026  ak1dnar 1,584
Recognized Expert Top Contributor
$sqlquery = "INSERT INTO $table VALUES('$id','$ name','$telepho ne','$birthday' )";
Did you create a $_POST Array to Collect the VALUES? I didn't see it on the Script.php.
No I did not use an array in this case, but instead submitted each value individually.
No I did not use an array in this case, but instead submitted each value individually.
Change your results line to - $results =mysql_query($sqlquery) or die("Error : ".mysql_error());
Also don't forget to make use of the poor man's debugger( echo, printf, or whatever you call those things in PHP).
ak1dnar 1,584
Recognized Expert Top Contributor
No I did not use an array in this case, but instead submitted each value individually.
Please you may better to read here.
Get the HTML form elements to variables. I think Now you might know what is $_POST.
- $name=$_POST['name'];
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$telephone=$_POST['telephone'];
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$birthday=$_POST['birthday'];
Pass The varibales to the MySQL Queary String. - $sqlquery = "INSERT INTO sample_table (name,telephone,birthday)
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VALUES (
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'$name', '$telephone', '$birthday'
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);";
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I didn't add the id formelement (may be it is the PK of your Table Structure). Actually value for that, has set up as Null. Still Only a String value. Its not same as the NULL on MySQL.
How is your Table Structure goes?
Your id column must be declared with auto_increment property. If u dont want to add id to your db by manually..
Kind Regards...
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