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Home/ Questions/Q 4325370
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Editorial Team
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Editorial Team
Asked: May 21, 20262026-05-21T09:15:27+00:00 2026-05-21T09:15:27+00:00

Ok, so im new in php and sql, and I have this form that

  • 0

Ok, so im new in php and sql, and I have this form that submits some names and cities into a database.

I managed to do it, but once a hit the submit button, i get an error:

“Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘1’ at line 1”

but, when i check in phpmyadmin, the new record is there!!, so im not sure what’s wrong, thats the problem.

this is the code:

<?php
   mysql_connect("localhost", "name", "pass") or die(mysql_error());
   echo "Connection to the server was successful!<br/>";

   mysql_select_db("db_name") or die(mysql_error());
   echo "Database was selected!<br/>";



    $resultComuna = mysql_query("SELECT idComuna, nombre FROM comuna ORDER BY nombre ASC");
    $resultGiro = mysql_query("SELECT idGiro, nombre FROM giro ORDER BY nombre ASC");

?>

<html>
<head>
    <title>TEST</title>
</head>
<body>
    <br/><br/>
    <form name="form" method="POST" action="test_action.php">
        <div align="center">

    <!--/////////////////  input nombre ////////////////////////  -->
            NOMBRE CLIENTE:
            <input name="nombreCliente" type="text" maxlength="30" size="40"></>

    <!-- ///////////////////////////////////////////////////////////// -->

    <!-- ////////////////////drop box para giro ///////////////////// -->
            GIRO:
            <select name="giro">
            <?php
            while($row = mysql_fetch_assoc($resultGiro)){
                echo "<option value=\"".$row['idGiro']."\">".$row['nombre']."</option><br/>";
            }
            ?>      
            </select>
    <!-- ///////////////////////////////////////////////////////////// -->

    <!-- ////////////// dropbox para comunas //////////////////////// -->
            COMUNA:
            <select name="comunas">
            <?php
            while($row = mysql_fetch_assoc($resultComuna)){

                echo "<option value=\"".$row['idComuna']."\">".$row['nombre']."</option><br/>";
            }
            ?>      
            </select>
    <!-- ////////////////////////////////////////////////////////////// -->

        <input type="submit" value="Ingresar"> </>

        </div>
    </form>

</body>
</html>

and the test_action.php is:

<?php
    $con = mysql_connect("localhost", "name", "pass");
    if (!$con)
      {
      die('Could not connect: ' . mysql_error());
      }

    mysql_select_db("data_base", $con);


$query = mysql_query("SELECT max(idNombre)+1 as id FROM nombre");
$row = mysql_fetch_array($query);
$idMax = $row['id'];


    $sql = mysql_query("INSERT INTO nombre VALUES ('".$idMax."','".$_POST['comunas']."',".$_POST['giro'].",'".$_POST['nombreCliente']."')");

    if (!mysql_query($sql,$con))
      {
      die('Error: ' . mysql_error());
      }
    echo "record added";

    mysql_close($con)
?>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-21T09:15:27+00:00Added an answer on May 21, 2026 at 9:15 am

    You’re inserting the ID in single quotes:

    $sql = mysql_query("INSERT INTO nombre VALUES ('".$idMax."','".$_POST['comunas']."',".$_POST['giro'].",'".$_POST['nombreCliente']."')");
    

    Can you provide the table structure? ID is an integer or a varchar there?

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