PHP doesn't insert into mysql

Question

I'am trying to do very simple php page and have some kind of problem here. the page works and everything but there is nothing coming to mysql. When I press submit it prints insert into DrinkHistory(CustomerId,DrinkId,SellerId) values('1','3','2')Your Data Inserted. But when I login to phpmyadmin there is nothing in DrinkHistory table.

the code is below

<?php
// Create connection
$con=mysqli_connect("localhost","******","*******","test");

// Check connection
if (mysqli_connect_errno($con))
  {
  echo "Mene kotii ku et mitää osaa: " . mysqli_connect_error();
  }

// Insert Data

@$a=$_POST['CustomerId'];
@$b=$_POST['DrinkId'];
@$c=$_POST['SellerId'];
if(@$_POST['submit'])
{
echo $s="insert into DrinkHistory(CustomerId,DrinkId,SellerId) values('$a','$b','$c')";
echo "Your Data Inserted";
mysql_query($s);
}

?>


<center>
<form method="post">
<table width="100%" height="245" border="1" bgcolor="#00CCFF">
<tr><td width="112" height="26">Asiakas</td>
<td width="100"><input type="radio" name="CustomerId" value="1"/>Jokirinne Niko</td>
</tr>

<tr><td rowspan="2">Juoma</td>
<form method)="post">
<td height="28"><input type="radio" name="DrinkId" value="2"/>Kalja</td>
<td height="28"><input type="radio" name="DrinkId" value="3"/>Lonkero</td>
<td height="28"><input type="radio" name="DrinkId" value="4"/>Siideri</td>
<td height="28"><input type="radio" name="DrinkId" value="5"/>Fisu</td>
<td height="28"><input type="radio" name="DrinkId" value="6"/>Tequila</td>
<td height="26"><input type="radio" name="DrinkId" value="7"/>MustikkaShotti</td>
</tr>
<td height="33"></tr>


<tr><td rowspan="3">Myyjä</td>
<td><input type="radio" name="SellerId" value="1"/>Niko Jokirinne</td>
<tr>
<td><input type="radio" name="SellerId" value="2"/>Tanya Lickorish</td>
<tr>

<tr><td height="62"><input type="submit" name="submit" value="Juo"/></td></tr>
</table>
</form>
</center>
</body>
</html>

mysqli_connect() + mysql_query() - error handling = problems ;-) — Álvaro González, May 09 '13 at 10:24
Can't see anything obvious wrong, can you have your table structure and the result of the $s echo? — We0, May 09 '13 at 10:26
Also, be careful because you have to
tags in your HTML, and the second one is wrong. — Juan Pablo Rinaldi, May 09 '13 at 10:26

score 0 · Answer 1 · edited May 23 '17 at 10:25

0

You have several database libraries to choose from but, once you decide, you need to stick to one. You cannot start a connection with the new mysqli extension:

$con=mysqli_connect("localhost","******","*******","test");

... and then try to query the database with the legacy deprecated functions:

mysql_query($s);

If you are not getting a blattant error message on screen you've probably failed to configure PHP to display error messages in your development box.That's something you need to fix before you go further; it's impossible to code without the aid of error messages. Here's a brief explanation.

Finally, you verify if your connection fails:

if (mysqli_connect_errno($con))

... but you don't test if your query fails.

edited May 23 '17 at 10:25

Community

1
1

answered May 09 '13 at 10:29

Álvaro González

142,137
41
261
360

After i wrote ini_set('display_errors', 1); error_reporting(~0); into code it says Warning: mysql_query() expects parameter 1 to be string, object given in /var/www/database.php on line 29 Warning: mysqli_error() expects exactly 1 parameter, 0 given in /var/www/database.php on line 32 Error inserting data: – user2365812 May 09 '13 at 10:37
I'm glad my second and third tips helped you. Now head to the first one ;-) – Álvaro González May 09 '13 at 10:47

score 0 · Accepted Answer · answered May 09 '13 at 10:30

0

You seem to be omitting the connection string (using mysql_query instead of mysqli_query) and error verification. Try this:

<?php
// Create connection
$con=mysqli_connect("localhost","******","*******","test");

// Check connection
if (mysqli_connect_errno($con))
  {
  echo "Mene kotii ku et mitää osaa: " . mysqli_connect_error();
  }

// Insert Data

@$a=$_POST['CustomerId'];
@$b=$_POST['DrinkId'];
@$c=$_POST['SellerId'];
if(@$_POST['submit']) {
    echo $s = "insert into DrinkHistory(CustomerId,DrinkId,SellerId) values('$a','$b','$c')";
    if (mysqli_query($con, $s)) {
        echo "Your Data Inserted";
    } else {
        echo "Error inserting data: ".mysqli_error();
    }
}

?>

answered May 09 '13 at 10:30

Francisco Zarabozo

3,676
2
28
54

now it says: "insert into DrinkHistory(CustomerId,DrinkId,SellerId) values('1','3','2')Error inserting data:" – user2365812 May 09 '13 at 10:32
Sorry, I forgot to change the function name. I just edited it to correct it. By the way: that code is 100% vulnerable to sql injection. You should always test the data you receive before trying to execute a query with interpolated variables received from a POST. – Francisco Zarabozo May 09 '13 at 10:34

PHP doesn't insert into mysql

2 Answers2