Question

在那里，我使用此代码使用HTML创建JSON表，在我的网络程序中，我使用JSON生成PHP，现在我需要传递{{ 1}}作为上述函数的参数，我该怎么做。

JSON

我可以使用<script type="text/javascript"> var data = [ { "UserID": 1, "UserName": "rooter", "Password": "12345", "Country": "UK", "Email": "sac@gmail.com" }, { "UserID": 2, "UserName": "binu", "Password": "123", "Country": "uk", "Email": "Binu@gmail.com" }, { "UserID": 3, "UserName": "cal", "Password": "123", "Country": "uk", "Email": "cal@gmail.com" }, { "UserID": 4, "UserName": "nera", "Password": "1234", "Country": "uk", "Email": "nera@gmail.com" }, { "UserID": 4, "UserName": "nera", "Password": "1234", "Country": "uk", "Email": "nera@gmail.com" } ]; $(document).ready(function () { var html = '<table class="table table-striped">'; html += '<tr>'; var flag = 0; $.each(data[0], function(index, value){ html += '<th>'+index+'</th>'; }); html += '</tr>'; $.each(data, function(index, value){ html += '<tr>'; $.each(value, function(index2, value2){ html += '<td>'+value2+'</td>'; }); html += '<tr>'; }); html += '</table>'; $('body').html(html); }); </script>

将JSON值作为参数传递给此函数

Answer 1

您可以使用php标记传递参数

   <?php
$data = '[
  {
    "UserID": 1,
    "UserName": "rooter",
    "Password": "12345",
    "Country": "UK",
    "Email": "sac@gmail.com"

  },
  {
    "UserID": 2,
    "UserName": "binu",
    "Password": "123",
    "Country": "uk",
    "Email": "Binu@gmail.com"
  },
  {
    "UserID": 3,
    "UserName": "cal",
    "Password": "123",
    "Country": "uk",
    "Email": "cal@gmail.com"
  },
  {
    "UserID": 4,
    "UserName": "nera",
    "Password": "1234",
    "Country": "uk",
    "Email": "nera@gmail.com"
  },
  {
    "UserID": 4,
    "UserName": "nera",
    "Password": "1234",
    "Country": "uk",
    "Email": "nera@gmail.com"
  }
]';


?>

<script>
    $(document).ready(function () {
    var html = '<table class="table table-striped">';
    html += '<tr>';
    var flag = 0;

    var data2   =   <?php echo $data; ?>;
    $.each(data2[0], function(index, value){
        html += '<th>'+index+'</th>';
    });
    html += '</tr>';
     $.each(data2, function(index, value){
         html += '<tr>';
        $.each(value, function(index2, value2){
            html += '<td>'+value2+'</td>';
        });
        html += '<tr>';
     });
     html += '</table>';
     $('body').html(html);
     console.log(html);
});
    </script>

Answer 2

我建议只在php中创建你的表，这样当它加载时它会在页面上，对于seo和用户体验来说会更好。

但是这就是当你从php加载页面时你可以打印javascript数据而不必发出ajax请求。

关键是<?= json_encode(array); ?> json encode会将php数组/关联数组解析为json对象。

我在其单独的脚本标记中执行此操作，以防出现解析或编辑语法错误。

<?php                                                                                                             
$data = [                                                                                                         
  [ "UserID" => 1, "UserName" => "rooter", "Password" => "12345", "Country" => "UK", "Email" => "sac@gmail.com" ],
  [ "UserID" => 2, "UserName" => "binu", "Password" => "123", "Country" => "uk", "Email" => "Binu@gmail.com" ],   
];                                                                                                                
?>                                                                                                                
<script>data = <?= json_encode($data, JSON_NUMERIC_CHECK); ?></script>                                                                
$(document).ready(function () {                                                                                   
    var html = '<table class="table table-striped">';                                                             
    html += '<tr>';                                                                                               
    var flag = 0;                                                                                                 
    $.each(data[0], function(index, value){                                                                       
        html += '<th>'+index+'</th>';                                                                             
    });                                                                                                           
    html += '</tr>';                                                                                              
     $.each(data, function(index, value){                                                                         
         html += '<tr>';                                                                                          
        $.each(value, function(index2, value2){                                                                   
            html += '<td>'+value2+'</td>';                                                                        
        });                                                                                                       
        html += '<tr>';                                                                                           
     });                                                                                                          
     html += '</table>';                                                                                          
     $('body').html(html);                                                                                        
});                                                                                                               
</script>

如何将参数传递给jQuery函数

2 个答案: