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$objConnect = mysql_connect("localhost","root","") or die(mysql_error());
$objDB = mysql_select_db("mydatabase1");
mysql_query("BEGIN");
$strSQL = "INSERT INTO customer ";

$strSQL .="(CustomerID,Name,Email,CountryCode,Budget,Used) ";
$strSQL .="VALUES ";
$strSQL .="('".$_POST["txtCustomerID"]."','".$_POST["txtName"]."','".$_POST["txtEmail"]."' ";
$strSQL .=",'".$_POST["txtCountryCode"]."','".$_POST["txtBudget"]."','".$_POST["txtUsed"]."') ";
//echo "first";
//echo $strSQL; 
$objQuery1 = mysql_query($strSQL);
$strSQL = "INSERT INTO customer ";
$strSQL .="(CustomerID,Name,Email,CountryCode,Budget,Used) ";
$strSQL .="VALUES ";
$strSQL .="('".$_POST["txtCustomerID"]."','".$_POST["txtName"]."','".$_POST["txtEmail"]."' ";
$strSQL .=",'".$_POST["txtCountryCode"]."','".$_POST["txtBudget"]."','".$_POST["txtUsed"]."') ";
//echo "second";
//echo $strSQL; exit;
$objQuery2 = mysql_query($strSQL);

*

if(($objQuery1) and ($objQuery2))
{
    mysql_query("COMMIT");
    echo "Save Done.";
}
else
{
    mysql_query("ROLLBACK");
    echo "Error Save [".$strSQL."]";
}
mysql_close($objConnect);

The Form:

<form action="php_mysql_transaction2.php" name="frmAdd" method="post">
<table width="600" border="1">
  <tr>
    <th width="91"> <div align="center">CustomerID </div></th>
    <th width="160"> <div align="center">Name </div></th>
    <th width="198"> <div align="center">Email </div></th>
    <th width="97"> <div align="center">CountryCode </div></th>
    <th width="70"> <div align="center">Budget </div></th>
    <th width="70"> <div align="center">Used </div></th>
  </tr>
  <tr>
    <td><div align="center"><input type="text" name="txtCustomerID" size="5"></div></td>
    <td><input type="text" name="txtName" size="20"></td>
    <td><input type="text" name="txtEmail" size="20"></td>
    <td><div align="center"><input type="text" name="txtCountryCode" size="2"></div></td>
    <td align="right"><input type="text" name="txtBudget" size="5"></td>
    <td align="right"><input type="text" name="txtUsed" size="5"></td>
  </tr>
  </table>
  <input type="submit" name="submit" value="submit">
</form>

if i submit the form data goes to customer table. it is not passing into if part:

if(($objQuery1) and ($objQuery2))
{
        mysql_query("COMMIT");
        echo "Save Done.";
}

if i submit the form data goes to customer table. it is not coming to if part. Any one check and tell i could not understand this

Martin Prikryl
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user3152777
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  • Have a read up about SQL injection prevention. Your script is wide open to it. http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php – bumperbox Jan 06 '14 at 06:43
  • Add or die after your mysql_query if you want to see what the error message is eg. mysql_query($strSQL) or die(mysql_error()); – bumperbox Jan 06 '14 at 06:44
  • after implementing die(mysql_error()); and i have assigned primary key for customer table cid. it is throwing "Duplicate entry '0' for key 'PRIMARY'" – user3152777 Jan 06 '14 at 07:21

0 Answers0