Question

我在投资表中有多行。我需要检索所有这些并使用json_encode将其发回这是我的代码

$project_code = $_GET['id'];    
$sql = "select * from investment where project_code = '$project_code'";
        $result = mysql_query($sql);
        while($x1 = mysql_fetch_array($result))
        {   $detail = $x1['detail'];
            $sector = $x1['sector'];
        }   
        echo json_encode(array(
            'a' => $detail,
            'b' => $sector
            ));

在javascript中我编写了以下脚本：

$('#myHref').change(function(){
        var value = $('#myHref').val();
        $.get('display_acc.php',{id:value},function(data)
         {
    data = JSON.parse(data); 
     $( '#a' ).html(data.a);
     $( '#b' ).html(data.b);
    });
       });

我的问题是它只返回最后一行的数据。我想收到所有行的数据。

我必须在id a and b显示{{1}} 我用表来显示它。 提前感谢

Answer 1

你需要循环中的数组

$project_code = $_GET['id'];    
$sql = "select * from investment where project_code = '$project_code'";
    $result = mysql_query($sql);
    $cnt=0;
    while($x1 = mysql_fetch_array($result))

    {   $detail[$cnt] = $x1['detail'];
        $sector[$cnt] = $x1['sector'];
        $cnt++;
    }   
    echo json_encode(array(
        'a' => $detail,
        'b' => $sector
        ));

并在javascript循环结果

Answer 2

这应该可以解决问题：

- OP解释后编辑 -

<强> PHP

$project_code = (int)$_GET['id'];    //better put (int) before to avoid SQL Injection
$sql = "select detail, sector from investment where project_code = '$project_code'"; //better to maintain if you set only the fields you will use
    $result = mysql_query($sql);
    $return = array('detail'=> array(), 'sector' => array()); //here you can put empty return array or set detail and sector key. Your choice will have impact in your front-end

    $return['my_extra_var' ] = $my_extra_var;
    $i = 0;
    while($x1 = mysql_fetch_array($result))
    {
        $return['table'][$i]['detail'] = $x1['detail'];
        $return['table'][$i]['sector'] = $x1['sector'];
        $i++;
    }   
    echo json_encode($return);

<强>的Javascript

... your ajax request ...
data = JSON.parse(data);
var tableData = data.table;
$.each(tableData, function(index, value) {
    $('#tableData').append('<tr><td>'+value.detail+'</td><td>'+value.sector+'</td></tr>'); //append row to tbody
});

<强> HTML

<table>
    <thead>
        <th>Details</th>
        <th>Sector of Investment</th>
    </thead>
    <tbody id="tableData">
    </tbody>
 </table>

Answer 3

看起来你没有积累这两个元素。尝试这样的事情：

array_push(your_array_name, $detail, $sector);

然后在完成所有操作后对数组进行编码。

如何在php中使用json_encode方法发送行？

3 个答案: