Question

以下javascript函数似乎只有当我有最后的confirm（）语句时才能使用，我最初在那里进行调试。当我把它拿出来时，delete_row.php似乎没有运行。另外，也许作为提示/旁注，当我在那里有确认声明时，它适用于除safari之外的所有浏览器......

function deleterow(form) {

    if (!confirm("Are you sure you want to delete?")) return false;

    var queryString = "?ID=";

    for (var i = 0; i < document.myForm.rows.length; i++) {
        if (document.myForm.rows[i].checked) {
            ID = document.myForm.rows[i].value;
            ID = ID.slice(0, -1);
            queryString += ID;
            queryString += "-";
        }
    }
    queryString = queryString.slice(0, -1);

    try{
        // Opera 8.0+, Firefox, Safari
        ajaxRequest = new XMLHttpRequest();
    } catch (e){
        // Internet Explorer Browsers
        try{
            ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
        } catch (e) {
            try{
                ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
            } catch (e){
                // Something went wrong
                alert("Your browser broke!");
                return false;
            }
        }
    }

    var ajaxRequest;  // The variable that makes Ajax possible!
    // Create a function that will receive data sent from the server
    ajaxRequest.onreadystatechange = function(){
        if(ajaxRequest.readyState == 4){
            var ajaxDisplay = document.getElementById('ajaxDiv');
            ajaxDisplay.innerHTML = ajaxRequest.responseText;
        }
    }

    ajaxRequest.open("GET", "delete_row.php" + queryString, true);
    ajaxRequest.send(null); 
    confirm('Delete successful!');
}

更新已解决

我通过以下js脚本更改

检查ajaxRequest的状态

ajaxRequest.onreadystatechange = function(){ // Create a function that will receive data sent from the server
    if(ajaxRequest.readyState == 4 && ajaxRequest.status == 200){
        var ajaxDisplay = document.getElementById('ajaxDiv');
        ajaxDisplay.innerHTML = ajaxRequest.responseText;
    }
    else{
        alert('An error has occurred making the request');
        return false;
    }
}

并注意到我从服务器返回0状态。一些谷歌搜索帮助我意识到错误在于我如何定义调用这些函数的按钮。

原始代码是：

<div style='float:left; margin-right:10px;'><input type="submit" onClick="deleterow(document.myForm)" VALUE="Delete ROWs"></div>

修复是：

<div style='float:left; margin-right:10px;'><input type="button" onClick="deleterow(document.myForm)" VALUE="Delete ROWs"></div>

（提交类型必须更改为按钮类型）

Answer 1

delete_row.php doesn't seem to run你有没有验证过这个问题，你能不能给if(ajaxRequest.readyState == 4){添加一个提醒我虽然没有表单的东西我尝试过你的JS但似乎工作正常，http://jsfiddle.net/6gjy6/你有没有得到任何JS Google Chromes控制台中的错误？您是否尝试使用appripriate url delete_row.php" + queryString,在浏览器上执行基本的“GET”请求，并查看服务器如何响应而不是AJAX调用。

试试这个：

var queryString = "?ID=";

for (var i = 0; i < document.myForm.rows.length; i++) {
    if (document.myForm.rows[i].checked) {
        ID = document.myForm.rows[i].value;
        ID = ID.slice(0, -1);
        queryString += ID;
        queryString += "-";
    }
}
queryString = queryString.slice(0, -1);
var ajaxRequest; 

try{
    // Opera 8.0+, Firefox, Safari
    ajaxRequest = new XMLHttpRequest();
} catch (e){
    // Internet Explorer Browsers
    try{
        ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
    } catch (e) {
        try{
            ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
        } catch (e){
            // Something went wrong
            alert("Your browser broke!");
            return false;
        }
    }
}

// Create a function that will receive data sent from the server
ajaxRequest.onreadystatechange = function(){
    if(ajaxRequest.readyState == 4){
        alert("received: " + ajaxRequest.responseText);
        var ajaxDisplay = document.getElementById('ajaxDiv');
        ajaxDisplay.innerHTML = ajaxRequest.responseText;
    }
}

ajaxRequest.open("GET", "delete_row.php" + queryString, true);
ajaxRequest.send(null);

Answer 2

在你的js put

的顶部保留你的confirm（）声明

window.alert = null ;

并尝试

让我检查一下

Answer 3

我很确定你应该在调用onreadystatechange之后设置open事件，否则清除处理程序。

javascript＆amp; ajax只能使用confirm（）

3 个答案: