Question

这是我的代码，它不起作用，它不会插入数据库。请帮我解决这个问题。

$orgexist = $_POST['orgName1'];

$_SESSION['id'] = $_POST['id'];
$orgid          = $_POST['id'];
$orgnme         = $_POST['orgName1'];
$orgdesc        = $_POST['orgDesc'];
$orgcat         = $_POST['cat'];
$orgdept        = $_POST['coldept'];
$orgvis         = $_POST['vision'];
$orgmis         = $_POST['mision'];
//get the value of category from database
//echo $orgdept;
$dept           = "SELECT `col_id`, `col_description` FROM `college` WHERE `col_description` = '$orgdept'";
$deptresult     = mysql_query($dept);
while ($rows = mysql_fetch_array($deptresult)) {
    $getcol = $rows['col_id'];
    //echo $getcol;
}

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

您正在使用deprected函数，请使用mysqli_query替换mysql_query。

有关更多参考资料，请参阅http://php.net/manual/en/function.mysql-query.php

如何在插入之前检查数据库中的现有数据

1 个答案: