我剛剛創建了一個將數據發送到php和php到數據庫的ajax函數。插入到dotp_task_log表中,工作正常。但進一步,當我需要添加數據到dotp_tasks添加到dotp_task_log後,它不會添加,我不能找到爲什麼...我得到的Gerror,這是我的PHP文件,它添加數據到數據庫。不向數據庫添加值
<?php
$currentUser = isset($_POST['currentUser']) ? $_POST['currentUser'] : '';
$currentTasken = isset($_POST['currentTasken']) ? $_POST['currentTasken'] : '';
$currentPercent = isset($_POST['currentPercent']) ? $_POST['currentPercent'] : '';
$con = mysql_connect("localhost", "root", "") or die(mysql_error());
if(!$con)
die('Could not connectzzz: ' . mysql_error());
mysql_select_db("foxi" , $con) or die ("could not load the database" . mysql_error());
$check = mysql_query("SELECT * FROM dotp_task_log");
$numrows = mysql_num_rows($check);
if($numrows >= 1)
{
//$pass = md5($pass);
$ins = mysql_query("INSERT INTO dotp_task_log (task_log_creator, task_log_Task) VALUES ('$currentUser' , '$currentTasken')") ;
if($ins)
{
$check = mysql_query("SELECT * FROM dotp_tasks");
$numrows = mysql_num_rows($check);
if($numrows > 1)
{
//$pass = md5($pass);
$inss = mysql_query("INSERT INTO dotp_tasks (task_percent_complete) VALUES ('$currentPercent') WHERE task_id='$currentTasken'") ;
if($inss)
{
die("Succesfully added Percent!");
}
else
{
die("GERROR");
}
}
else
{
die("Log already exists!");
}
}
else
{
die("ERROR");
}
}
else
{
die("Log already exists!");
}
?>
錯誤檢查。 [1](http://php.net/manual/en/function.error-reporting.php) - [2](http://php.net/manual/en/function.mysql-error.php) –
你什麼意思? – TorresAlGrande
刷新我的評論,他們是超鏈接 –