Question

我成功从mysql数据库获取数据，直到我尝试删除表。该表下降，但我无法弄清楚如何保持javascript ajax数据库访问声明错误。

如果没有要返回的数据，我怎么能回来？

以下php例程删除订单表中的一行，然后删除名为$ tableNameInput的表。检查数据库显示两个操作都已成功执行但javascript ajax错误处理因x.status = 0而被触发。我只需要一种方法来告诉javascript ajax它是否真的完整。

function dropCartTable($tableNameInput){/* drops the cart table and its order entry*/
    AnnLog("dropCartTable: ",$tableNameInput);
    // first drop the reference in the orders table
    $tableName = "orders";
    $query = "DELETE FROM $tableName WHERE cartDB ='$tableNameInput'";
    $result = doQuery($tableName,$query, "Could not drop orders row");
    if($result){
        // now drop this cart table
        $sql = "DROP TABLE IF EXISTS $tableNameInput";
    }
    //********do query  just does the query and logs.   it does not return if it dies.
    $result = doQuery($tableName, $sql, "Can not drop csrt table");
    if($result){

        /****** when we get to here we have successfully dropped the items. below, I tried to generate a response and it did not work.*/

        $arr = array('drop cart' => 0);
        $j = json_encode ($arr);
        echo \json_encode($j);
        AnnLog("echoed ",$j);/* this writes my logs which verify that all steps until the return are ok.*/

    }
}

提前感谢您的帮助。

Answer 1

function dropCartTable($tableNameInput){/* drops the cart table and its order entry*/
AnnLog("dropCartTable: ",$tableNameInput);
// first drop the reference in the orders table
$tableName = "orders";
$query = "DELETE FROM $tableName WHERE cartDB ='$tableNameInput'";
$result = doQuery($tableName,$query, "Could not drop orders row");
if($result)
{
         // now drop this cart table
        $sql = "DROP TABLE IF EXISTS $tableNameInput";
$arr = array('drop cart' => 0);
     $j = json_encode ($arr);
    echo \json_encode($j);
AnnLog("echoed ",$j);/* this writes my logs which verify that all steps until the return are ok.*/

}
else
{
$arr = array("couldn't drop cart" => 0);
         $j = json_encode ($arr);
        echo \json_encode($j);
    AnnLog("echoed ",$j);/* .*/
}

问题是你正在丢弃表并返回结果if 它的真实和无视的错误

Answer 2

问题不在于json回归。这是因为我没有成功回调ajax调用。

php成功删除表后返回错误

2 个答案: