如何使用变量设置 Ajax Url

Question

我想在ajax进程中向url添加一些字符串，但每次访问该函数时都失败。当我提交表单时，它只能访问这个 url 'Master_data/'

var jenis_submit = document.getElementById('jenis_submit');

var submit_url = '';

if (jenis_submit == '1') {

    submit_url = "func_save_instansi";

} else if (jenis_submit == '2') {

    submit_url = "func_update_instansi";

}

$.ajax({

    url: '<?php echo base_url();?>Master_data/' + submit_url,

    method: 'POST',

    data: dataKirim,

    contentType: false,

    cache: false,

    processData: false,

    success: function(data) {

        // console.log(data);

        $('#notif').html(data);

    }

});

Answer 1

首先，像这样使用javascript通过id获取值，//var number = document.getElementById("number").value; //in your code `.value` is missingvar jenis_submit = document.getElementById('jenis_submit').value;var submit_url = '';if (jenis_submit == '1') {&nbsp; &nbsp; submit_url = "func_save_instansi";} else if (jenis_submit == '2') {&nbsp; &nbsp; submit_url = "func_update_instansi";}&nbsp;然后用这个替换你的代码url: "<?php echo base_url('master_data/'); ?>" + submit_url,

Answer 2

试试这个：var jenis_submit = document.getElementById('jenis_submit');var submit_url&nbsp; &nbsp;= '';if (jenis_submit == '1') {&nbsp; &nbsp; submit_url = "func_save_instansi";} else if(jenis_submit == '2') {&nbsp; &nbsp; submit_url = "func_update_instansi";}var post_url = '<?php echo base_url('Master_data'); ?>' + '/' + submit_url;$.ajax({&nbsp; &nbsp; &nbsp;type: "POST",&nbsp; &nbsp; &nbsp;url: post_url,&nbsp; &nbsp; &nbsp;data: dataKirim,&nbsp; &nbsp; &nbsp;beforeSend: function() {&nbsp; &nbsp; &nbsp; &nbsp;// you can show a preloader here&nbsp; &nbsp; &nbsp;}&nbsp;}).done(function(data) {&nbsp; &nbsp; &nbsp;// hide the preloader&nbsp; &nbsp; &nbsp;// console.log(data);&nbsp; &nbsp; &nbsp;$('#notif').html(data);&nbsp;}).fail(function() {&nbsp;&nbsp; &nbsp;// hide the preloader&nbsp; &nbsp;// show an error message&nbsp;});

Answer 3

只需将 php 值分配给 js 变量并连接两者var&nbsp;url&nbsp;=&nbsp;'<?php&nbsp;echo&nbsp;base_url();?>Master_data/';然后在ajax中url:&nbsp;url&nbsp;+&nbsp;submit_url,更新：我错过了您代码中的另一个错误var&nbsp;number&nbsp;=&nbsp;document.getElementById("number").value;&nbsp;//in&nbsp;your&nbsp;code&nbsp;`.value`&nbsp;is&nbsp;missing如上面（M.Hemant）的回答所述。