如何在插入php之前检查数据库中是否存在数据

Question

$sqlorg = mysql_query("SELECT * FROM `organization`");
while($orgrows = mysql_fetch_array($sqlorg)) {
    //$dborgid = $orgrows['org_id'];
    $dborgnme = $orgrows['org_name'];
}

if ($dborgnme == $orgexist) {
    echo "<script type='text/javascript'>
            alert('Organization Name Already Used by other Organization');
            history.back();
        </script>";
} else {
    $orginsrt = mysql_query("INSERT INTO `organization`(`org_id`,`org_name`,`org_desc`,`category`,`vision`,`mission`,`col_id`,`image`) VALUES ('$orgid','$orgexist','$orgdesc','$orgcat','$orgvis','$orgmis','$getcol','$image')");

    echo "<script type='text/javascript'>
        alert('Proceed to next Step');</script>";
        //require ('orgsignup.php');
        header ('Location:orgsignup2.php');
    //echo "Not in the Record";

}

Answer 1

There are multiple issues with this question, and as such can't easily be answered - I'm writing this "answer" as a quick guide to Kio Rii and to get more information:

1) Don't use MySQL, use MySQLi for procedural DB/PHP interactions.

2)

while($orgrows = mysql_fetch_array($sqlorg)) {
    //$dborgid = $orgrows['org_id'];
    $dborgnme = $orgrows['org_name'];
}

The value $dborgnme will only ever hold the final value from all the rows fetched from the database. Consider reformatting the While statement to wrap outside the following if(){} ... else{}

3) Add some context and information to your question - what are you checking for? Where do values such as $orgexist come from? What events do you want to occur if a data exists or what happens if a data doesn't exist?

4) If you're only checking the name, you can better do it with a better SELECT statement as the current select is grabbing all the MySQL rows, so try

take :

$sqlorg = mysql_query("SELECT * FROM `organization`");

and turn it into

$sqlorg = mysql_query("SELECT * FROM `organization` WHERE org_name LIKE '%".$orgexist."%' ");

which will only return you rows if the names are very similar. You can then use the output of this (count of rows returned) to carry on the script logic.

5) Yes, my solution in part 4 is quick and dirty, and PDO or MySQLi prepared statements are much, MUCH better and more secure than old MySQL and variable injection into the SQL statement.

Additional debugging:

you would probably find this very useful:

$orginsrt = mysql_query("INSERT INTO `organization`(`org_id`,`org_name`,`org_desc`,`category`,`vision`,`mission`,`col_id`,`image`) VALUES ('$orgid','$orgexist','$orgdesc','$orgcat','$orgvis','$orgmis','$getcol','$image')") or die("insert fail: ".mysql_error());

This will tell you why an insert failed, if it did fail.