Insert data into multiple tables using one form [duplicate]

Question

I am trying to insert data in 2 tables using one form.

This is my form

<form action="don.php" method="post">
<tr><td>
    <p>
        <label for="nume">Nume:</label></td>
        <td><input type="text" name="nume" id="nume" autocomplete="off"></td>
    </p>
    </tr>
    <tr><td>
    <p>
        <label for="prenume">Prenume:</label></td>
       <td> <input type="text" name="prenume" id="prenume" autocomplete="off"></td>
    </p></tr>
    <tr><td>
    <p>
        <label for="grupa">Numar telefon:</label></td>
        <td><input type="text" name="numar" id="numar" autocomplete="off"></td>
    </p></tr>
    <tr><td>
    <p>
        <label for="grupa">Suma:</label></td>
        <td><input type="text" name="suma" id="suma" autocomplete="off"></td>
    </p></tr>
    <tr><td>
    <p>
        <label for="grupa">Data:</label></td>
        <td><input type="text" name="data" id="data" autocomplete="off"></td>
    </p></tr>
    <tr><td>
    <p>
        <label for="grupa">IBAN:</label></td>
        <td><input type="text" name="iban" id="iban" autocomplete="off"></td>
    </p></tr>
    <tr><td><input type="submit" name="submit" value="Donează" onclick="alert('Operatiune finalizata cu succes. Va multumim!')"></td>
</form>

And this is my PHP code

<?php
if(isset($_POST['submit'])){
    $con=mysql_connect("localhost","root","");
    if(!$con)
    {
        die("Nu se poate face conexiunea la baza de date" . mysql_error());
    }

    mysql_select_db("laborator",$con);
    $sql="INSERT INTO donator (nume, prenume, numar_telefon) VALUES ('$_POST[nume]','$_POST[prenume]','$_POST[numar]')";
    $sql="INSERT INTO donatie (suma, data_donatie, IBAN) VALUES ('$_POST[suma]','$_POST[data]','$_POST[iban]')";
    mysql_query($sql,$con);
    mysql_close($con);
}
?>

When I press the submit button it shows me the alert that my dates were inserted, but only the second INSERT works. Table donator is empty. What should I do to fix this problem?

mysqli's multi query works well php.net/manual/en/mysqli.multi-query.php — Funk Forty Niner
– Funk Forty Niner, Commented May 26, 2015 at 18:45
Important: Validate your input: dreamhost.com/blog/2013/05/22/… — Scoutman
– Scoutman, Commented May 26, 2015 at 19:54

Scoutman · Accepted Answer · 2015-05-26 18:59:07Z

17

You must call mysql_query() for every query.

$sql1 = "INSERT INTO donator ...";
$sql2 = "INSERT INTO donatie ...";
mysql_query($sql1, $con);
mysql_query($sql2, $con);

Important

mysql_query() is deprecated! Please use mysqli_query() http://php.net/manual/de/book.mysqli.php

You can also use mysqli_multi_query() http://php.net/manual/de/mysqli.multi-query.php

$query = "INSERT INTO donator ...; INSERT INTO donatie ...;";
mysqli_multi_query($link, $query);

edited May 26, 2015 at 18:59

answered May 26, 2015 at 18:52

Scoutman

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Comments

Community · Accepted Answer · 2017-05-23 10:33:01Z

0

You run your queries with the same variable $sql. You should call them differentely like that and call them after.

$sql="INSERT INTO donator (nume, prenume, numar_telefon) VALUES ('mysql_real_escape_string($_POST[nume])','mysql_real_escape_string($_POST[prenume])','mysql_real_escape_string($_POST[numar])')";
$sql2="INSERT INTO donatie (suma, data_donatie, IBAN) VALUES ('mysql_real_escape_string($_POST[suma])','mysql_real_escape_string($_POST[data])','mysql_real_escape_string($_POST[iban])')";

Otherwise, you must change your post values by protecting them. Refer here

Then, you have to switch to new mysql syntax for PHP, like mysqli or PDO. THe mysql syntax you are using is deprecated.

edited May 23, 2017 at 10:33

CommunityBot

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answered May 26, 2015 at 18:45

Ɛɔıs3

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2 Comments

Eleanordum Over a year ago

I already tried to call them differently. And it doesn't work neither with sql . =...

Ɛɔıs3 Over a year ago

Because you have to call them with different queries

Shane Lessard · Accepted Answer · 2015-05-26 18:54:00Z

0

You're overwriting your $sql variable. Save them as separate variables, or run the first query before declaring the second one.

That said, you should REALLY consider both switching two either PDO or mysqli for your queries rather than the mysql extension and also looking into prepared statements and properly sanitizing strings, because you're very vulnerable to sql injections.

answered May 26, 2015 at 18:54

Shane Lessard

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Comments

Vidya · Accepted Answer · 2018-02-15 09:49:41Z

0

Use like this It will work perfectly.

<?php
$GLOBALS['server']="localhost";
$GLOBALS['username']="root";
$GLOBALS['password']="*****";
$GLOBALS['database']="performance";

$GLOBALS['conn']= mysqli_connect($server,$username,$password,$database);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
//echo "Connected successfully";    

$GLOBALS['db'] = mysqli_connect($server,$username,$password,$database);

$sql="INSERT INTO animals (id,name) VALUES('2211','vidya');";
$sql .="INSERT INTO birds (id,fame) VALUES('2211','viddi');";
//mysqli_query($conn,$sql);
//mysqli_query($conn,$sql1);
if (mysqli_multi_query($GLOBALS['db'], $sql)) {
    echo "New records created successfully";
} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}

mysqli_close($conn);
?>

answered Feb 15, 2018 at 9:49

Vidya

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1 Comment

Exception_al Over a year ago

Welcome to SO. It would be nice if you could also explain as to why your suggestion should work and where the OP might've made a mistake.

thisal selaka · Accepted Answer · 2018-11-04 07:40:42Z

This is a example of one form data insert for a multiple tables.

Here html form has one submit button. "ok" name is the submit button name.

<?php
ob_start();
session_start();
$con=mysqli_connect("localhost","root","","cultureframework");
if(mysqli_connect_errno())
{
    echo "Failed to connect to MySql:".mysqli_connect_error();
}
?>
<?php
if(isset($_POST['ok']))
{
    $ip=$_POST['ip'];
    $date=$_POST['date'];
    $gender=$_POST['gender'];
    $age=$_POST['age'];
    $tenure=$_POST['tenure'];
    //-------------------------------------
    $caring_past_1=$_POST['caring_past_1'];
    $caring_present_1=$_POST['caring_present_1'];
    $caring_future_1=$_POST['caring_future_1'];
    $caring_past_2=$_POST['caring_past_2'];
    $caring_present_2=$_POST['caring_present_2'];
    $caring_future_2=$_POST['caring_future_2'];

    //-------------------------------------

        $insert="INSERT INTO `generalinfo`(`ip`,`datez`,`gender`,`age`,`tenure`) VALUES('$ip','$date','$gender','$age','$tenure')";
        $query=mysqli_query($con,$insert);

        $ins="INSERT INTO `caring`(`caring_past_1`,`caring_present_1`,`caring_future_1`,`caring_past_2`,
        `caring_present_2`,`caring_future_2`) VALUES('$caring_past_1','$caring_present_1','$caring_future_1','$caring_past_2','$caring_present_2','$caring_future_2')";
        $quy=mysqli_query($con,$ins);   
}

?>

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