Question

我想更新mysql表但只有一次。如果你想再次更新它，它应该给出消息＆＃34; ID已经在使用中，请登录＆＃34;

<?php 
If(isset($_POST['login'])){
$Pin=$_GET['pin'];
$ID =$_POST['ID'];
if($Pin!=''){
$result = mysql_query("SELECT * FROM pin WHERE Pin  = '$Pin'");
$test = mysql_fetch_array($result);

mysql_query("UPDATE pin SET appid ='$num' WHERE Pin= '$Pin'")
            or die(mysql_error()); 

header("location:compet_applicant.php");
}
}

?>

Answer 1

我认为您应该首先在field表格中再添加一个number_update，如：pin。默认值为0 第一次更新时，它的值为1，您可以检查该值，如果是1，则会提醒ID Already in Use, Pls login。如果是0，则允许更新

$result = mysql_query("SELECT * FROM pin WHERE Pin  = '$Pin'");
$test = array();
while ($row = mysql_fetch_array($result)) {
    $test[] = array_map('utf8_encode', $row);
}

if($test["number_update"] == 1) { //Checking already updated
    //Notify user that they have already updated
} else {
    mysql_query("UPDATE pin SET appid ='$num' WHERE Pin= '$Pin'")
        or die(mysql_error());
}

Answer 2

<?php
if (isset($_POST['login'])) {
    $Pin = $_GET['pin'];
    $ID  = $_POST['ID'];
    if ($Pin != '') {
        $result = mysql_query("SELECT * FROM pin WHERE Pin  = '$Pin'");
        $test   = array();
        while ($row = mysql_fetch_array($result)) {
            $test[] = array_map('utf8_encode', $row);
        }

        if ($test["number_update"] == 1) { //Checking already updated
            //Notify user that they have already updated
        } else {
            mysql_query("UPDATE pin SET appid ='$num',number_update = 1 WHERE Pin= '$Pin'") or die(mysql_error());
        }

        header("location:compet_applicant.php");
    }
}
?>

按照Neo先生的建议在数据库表中再添加一个字段“number_update”，并用上面的代码替换您的代码。

Answer 3

应该有一个记录表的每个更新的日志表。如果找到记录，那么系统应阻止该特定记录的进一步更新。

    <?php 
    If(isset($_POST['login'])){
    $Pin=$_GET['pin'];
    $ID =$_POST['ID'];
    if($Pin!=''){
    $update_count = mysql_query("SELECT * FROM Audit_Table_Update WHERE Field_name  = 'Pin' and field_value='$pin');

if(!mysql_num_rows($update_count)){

    $result = mysql_query("SELECT * FROM pin WHERE Pin  = '$Pin'");
    $test = mysql_fetch_array($result);

    mysql_query("UPDATE pin SET appid ='$num' WHERE Pin= '$Pin'")
                or die(mysql_error()); 

    mysql_query("INSERT INTO AUDIT_TABLE_UPDATES (TABLE_NAME, FIELD_NAME, FIELD_VALUE,TIME_UPDATED) VALUES('Pin','appid',$num,systemtime)")
                or die(mysql_error()); 

    header("location:compet_applicant.php");
    } // end if update count
    }
    }
    ?>

php mysql更新表一次

3 个答案: