Javascript php在成功删除我的表后返回错误

在尝试删除表之前，我一直在成功地从mysql数据库获取数据。该表将被删除，但我不知道如何防止javascript ajax数据库访问声明错误

没有数据可返回时，如何返回ok

下面的php例程删除order表中的一行，然后删除名为$tableNameInput的表。对数据库的检查表明，这两个操作都已成功执行，但javascript ajax错误处理会在x.status=0时中断。我只需要一种方法来告诉javascript ajax它真的完成了

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.*/

    }
}

提前感谢你的帮助

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);/* .*/
}

问题是您正在删除表并在 是真是假

---

问题不在json返回中。这是由于我没有成功回调ajax调用造成的

你是双重编码你的JSON为什么不只是

echo JSON_encode（$arr）？不知怎的，上面的一些行被删除了。实际上有两个查询，一个查询到订单表，另一个查询到购物车表。在这两种情况下，我都调用一个名为doquery的函数，该函数处理查询并将日志记录到我的日志文件中。如果查询失败，日志文件将更新，进程将终止。if（$结果是rednundant。我的问题是json_编码没有返回到调用的rountine。我得到错误号0。格式错误。我们如何关闭它？