jQuery Validation Plugin - 成功验证后无法链接函数[重复]

Posted 2023-05-09

【中文标题】jQuery Validation Plugin - 成功验证后无法链接函数[重复]

【英文标题】：jQuery Validation Plugin - can't chain function after successful validation [duplicate]

【发布时间】：2018-01-23 20:09:42

【问题描述】：

所以我想做的就是这部分代码：

$(document).ready(function ()  //predaj formu i automatski crtaj graf
    $('#myform').on('submit', function (e) 
      e.preventDefault();

      $.ajax(
        type: 'post',
        url: 'upit.php',
        data: $('#myform').serialize(),
        success: function () 
            var dataPoints = [];
            $.getJSON("rez.json", function(data)  //uzmi JSON za tocke grafa
                $.each(data, function(key, value)
                    dataPoints.push(x: value[0], y: parseInt(value[1]));
                );

                var chart = new CanvasJS.Chart("chartContainer",
                    zoomEnabled: true,
                    animationEnabled: false,
                    axisY: 
                        title: "Power received"
                    ,
                    axisX: 
                        title: "Distance"
                    ,
                    data: [
                        type: "line",
                        dataPoints : dataPoints,
                    ]
                );
                chart.render();
            );

            $.ajax( //vrati rezultat
                url:"novi.json",
                success:function(result)
                    $("#disabledInput").val(result);
                
            );        
        
    );
);
);

执行

通过这部分代码验证成功提交后：

$(document).ready(function () 

$('#myform').validate( // initialize the plugin
    rules: 
        n1: 
            required: true,
            email: true
        ,
        n2: 
            required: true,
            minlength: 5
        
    ,
    errorPlacement: function(error, element) 
        error.appendTo('#nameError');
    ,

    submitHandler: function (form)  // for demo
        alert('valid form submitted'); // for demo
        return false; // for demo
    
);

);

我只是无法正确链接它，我知道这两个是分开的脚本，所以我需要它们以某种方式链接。谢谢！

此外，我需要将 ajax 提交（第一个代码）代码分开，因为我想用许多其他函数调用它（播放是添加一个滑块，所以我在每次更改滑块等时提交它......）但我'不知道如何

编辑：我做到了，这是答案

$(document).ready(function () 

$('#myform').validate( // initialize the plugin
rules: 
    n1: 
        required: true,

    ,
    n2: 
        required: true,

    
,
errorPlacement: function(error, element) 
    error.appendTo('#nameError');
,

submitHandler: function (form) 


      $.ajax(
        type: 'post',
        url: 'upit.php',
        data: $('#myform').serialize(),
        success: function()
            var dataPoints = [];
            $.getJSON("rez.json", function(data)  //uzmi JSON za tocke 
grafa
                $.each(data, function(key, value)
                    dataPoints.push(x: value[0], y: parseInt(value[1]));
                );

                var chart = new CanvasJS.Chart("chartContainer",
                    zoomEnabled: true,
                    animationEnabled: false,
                    axisY: 
                        title: "Power received"
                    ,
                    axisX: 
                        title: "Distance"
                    ,
                    data: [
                        type: "line",
                        dataPoints : dataPoints,
                    ]
                );
                chart.render();
            );


        
    );

);

);

【问题讨论】：

如果您自己解决了问题，请将解决方案发布在答案中，而不是问题中。

【参考方案1】：

我做到了，这是答案

$(document).ready(function () 

$('#myform').validate( // initialize the plugin
rules: 
    n1: 
        required: true,

    ,
    n2: 
        required: true,

    
,
errorPlacement: function(error, element) 
    error.appendTo('#nameError');
,

submitHandler: function (form) 


      $.ajax(
        type: 'post',
        url: 'upit.php',
        data: $('#myform').serialize(),
        success: function()
            var dataPoints = [];
            $.getJSON("rez.json", function(data)  //uzmi JSON za tocke 
grafa
                $.each(data, function(key, value)
                    dataPoints.push(x: value[0], y: parseInt(value[1]));
                );

                var chart = new CanvasJS.Chart("chartContainer",
                    zoomEnabled: true,
                    animationEnabled: false,
                    axisY: 
                        title: "Power received"
                    ,
                    axisX: 
                        title: "Distance"
                    ,
                    data: [
                        type: "line",
                        dataPoints : dataPoints,
                    ]
                );
                chart.render();
            );


        
    );

);

);