Question

为什么我的jquery中出现错误，即使我的记录插入数据库？我在这段代码中添加了一些更改，我将其转换为mysqli函数。但问题是错误信息出来了吗？那是为什么？

错误消息来自我的脚本

这是我的代码

脚本

$(document).ready(function(){

    $("#save").click(function(){
        ajax("save");
    });

    $("#add_new").click(function(){
        $(".entry-form").fadeIn("fast");    
    });

    $("#close").click(function(){
        $(".entry-form").fadeOut("fast");   
    });

    $("#cancel").click(function(){
        $(".entry-form").fadeOut("fast");   
    });

    $(".del").live("click",function(){
        if(confirm("Do you really want to delete this record ?")){
            ajax("delete",$(this).attr("id"));
        }
    });

    function ajax(action,id){
        if(action =="save")
            data = $("#userinfo").serialize()+"&action="+action;
        else if(action == "delete"){
            data = "action="+action+"&item_id="+id;
        }

        $.ajax({
            type: "POST", 
            url: "ajax.php", 
            data : data,
            dataType: "json",
            success: function(response){
                if(response.success == "1"){
                    if(action == "save"){
                        $(".entry-form").fadeOut("fast",function(){
                            $(".table-list").append("<tr><td>"+response.cat_name+"</td><td>"+response.cat_code+"</td><td>"+response.letter+"</td><td><a href='#' id='"+response.row_id+"' class='del'>Delete</a></td></tr>");   
                            $(".table-list tr:last").effect("highlight", {
                                color: '#4BADF5'
                            }, 0000);
                        }); 
                        $(".entry-form input[type='text']").each(function(){
                            $(this).val("");
                        });                     
                    }else if(action == "delete"){
                        var row_id = response.item_id;
                        $("a[id='"+row_id+"']").closest("tr").effect("highlight", {
                            color: '#4BADF5'
                        }, 0000);
                        $("a[id='"+row_id+"']").closest("tr").fadeOut();
                    }
                }else{
                    alert("unexpected error occured, Please check your database connection");
                }
            },
            error: function(res){
                alert("Unexpected error! Try again.");
            }
        });
    }
});

AJAX

<?php
error_reporting(0);
$mysqli = new mysqli("localhost", "root", "", "2015");
if(isset($_POST) && count($_POST)){
    $cat_name = $mysqli->real_escape_string($_POST["cat_name"]);
    $cat_code = $mysqli->real_escape_string($_POST["cat_code"]);
    $letter = $mysqli->real_escape_string($_POST["letter"]);
    //$phone = mysql_real_escape_string($_POST['phone']);
    $item_id = $_POST['item_id'];
    $action = $_POST['action'];

    if($action == "save"){
        $result = $mysqli->query("insert into category values('','".$cat_name."','".$cat_code."','".$letter."')");
        $lid = mysqli_insert_id();
        if($lid){
            echo json_encode(
                array(
                "success" => "1",
                "row_id" => $lid,
                "cat_name" => htmlentities($cat_name),
                "cat_code" => htmlentities($cat_code),
                "letter" => htmlentities($letter),
                //"phone"=>$unique_id,
                )
            );
        }else{
            echo json_encode(array("success" => "0"));
        }
    }
    else if($action == "delete"){
        //echo "delete from info where id = '".$item_id."'";
        $res = $mysqli->query("delete from category where id = '".$item_id."'");
        if($res){
            echo json_encode(array( "success" => "1","item_id" => $item_id));
        }else{
            echo json_encode(array("success" => "0"));
        }
    }
}else{
    echo json_encode(array("success" => "0"));
}
?>

Answer 1

成功块中返回的Reponse为零，因为此处的语句$lid = mysqli_insert_id();没有返回任何内容。

将$lid = mysqli_insert_id();替换为$mysqli->insert_id，因为这是获取最近添加/插入行的ID的正确语法。

快乐编码：）

Answer 2

我明白了。 mysqli_insert_id（）需要连接数据库。

应该是这样的

mysqli_insert_id($mysqli);

即使记录正确插入也会出错

2 个答案: