JSON 值转换为 HTML 表

发布于 2024-11-03 12:26:31 字数 605 浏览 1 评论 0原文

我在同一文件夹中有 table.html、data.php 和 json.csv 。

data.php 正在执行 fopen("json.csv","r") 来读取 json.csv。

如何将 JSON 对象显示为 table.html 中的表格？

<html>
<head>
<script type="text/javascript" src="jquery.js">
function display(){
 $.post('data.php',function(data){
 $("#txtJSON").html(data); 
 });
  }

  </script>
  </head>
  <body onload="display()">
  <table id="#jobtable">
  <input type="text" id="txtJSON" />
  </table>
  </body>
  </html>

原文

I have a table.html, data.php and json.csv within the same folder.

data.php is doing fopen("json.csv","r") to read from json.csv.

How can I display the JSON objects as a table within table.html?

<html>
<head>
<script type="text/javascript" src="jquery.js">
function display(){
 $.post('data.php',function(data){
 $("#txtJSON").html(data); 
 });
  }

  </script>
  </head>
  <body onload="display()">
  <table id="#jobtable">
  <input type="text" id="txtJSON" />
  </table>
  </body>
  </html>

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晨敛清荷 2024-11-10 12:26:31

您只需要 json_decode 和一些迭代

http://php.net/manual/ en/function.json-decode.php

我将向您展示一种非常简单的 PHP 方法。您没有透露有关 JSON 的任何信息，因此我假设它是扁平的（不是嵌套的）。我想你可能也想改变客户端的东西。将 txtJSON 放入一个空的 div 中，其中将填充 PHP 的输出。

$file = $fopen("json.csv", 'r');
$fileText = fread($file, filesize($file));
$arr = json_decode($fileText, true); //i prefer associative array in this context

echo "<table>";
foreach($arr as $k=>$v)
    echo "<tr><td>$k</td><td>$v</td></tr>";
echo "</table>";

如果您的 JSON 不平坦且不简单，您希望生成的表是什么样子？

You just need json_decode and some iteration

http://php.net/manual/en/function.json-decode.php

I'll show you a very simplistic way to do the PHP. You don't tell anything about the JSON, so I'm assuming its flat (not nested). I think you might want to alter the client side stuff too. Make txtJSON into an empty div, which will be filled with the output from the PHP.

$file = $fopen("json.csv", 'r');
$fileText = fread($file, filesize($file));
$arr = json_decode($fileText, true); //i prefer associative array in this context

echo "<table>";
foreach($arr as $k=>$v)
    echo "<tr><td>$k</td><td>$v</td></tr>";
echo "</table>";

If your JSON is not flat and simple, what would you like the resulting table to look like?

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浮萍、无处依 2024-11-10 12:26:31

这里我将 json 数据放入 $json 变量中，然后使用 foreach 循环从 json 数据创建表。

foreach ($data as $key => $jsons) {
 $table ='<table class="'.$jsons['class'].'" border="1">';
 foreach ($jsons as $rkey => $rvalue) {
    if($rkey=='head')
    {
        $table.='<tr>';
        foreach($rvalue as $rvv)
        {
            $table.='<th>'.$rvv.'</th>';
        }
        $table.='</tr>';
    }else
    if($rkey=='rows')
    {
        foreach($rvalue as $rvv)
        {
            $table.='<tr>';
            foreach($rvv as $rv)
            {
                $table.='<td>'.$rv.'</td>';
            }
            $table.='</tr>';
        }
    }
 }
}
echo $table;
?>

我参考了 http://www.tutsway.com/create-table-来自-json.php

Here I have taken json data in to $json variable, and then use foreach loop to create table from json data.

foreach ($data as $key => $jsons) {
 $table ='<table class="'.$jsons['class'].'" border="1">';
 foreach ($jsons as $rkey => $rvalue) {
    if($rkey=='head')
    {
        $table.='<tr>';
        foreach($rvalue as $rvv)
        {
            $table.='<th>'.$rvv.'</th>';
        }
        $table.='</tr>';
    }else
    if($rkey=='rows')
    {
        foreach($rvalue as $rvv)
        {
            $table.='<tr>';
            foreach($rvv as $rv)
            {
                $table.='<td>'.$rv.'</td>';
            }
            $table.='</tr>';
        }
    }
 }
}
echo $table;
?>

I have taken reference from http://www.tutsway.com/create-table-from-json.php

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