向JSON编码的数组添加其他对象

继续往上推 $data 数组。

$json = array();

    while($row = mysql_fetch_assoc($sth)) {
        $json['name'] = $row['name'];
        $json['id'] = $row['id'];
        $data[] = $json;
    }

$custom = array('name'=>'foo', 'id' => 'bar');
$data[] = $custom;

最后,做你的工作 json_encode . 假设您不是指将它与多个ajax调用合并到JS本身中。

如果您有单独的脚本,请将它们合并到一个php页面中。

$temp = json_encode($json);  //$json={"var1":"value1","var2":"value2"}   
$temp=substr($temp,0,-1);
$temp.=',"variable":"'.$value.'"}';

您可以编辑JSON(文本),但是在编码之前修改数组要容易得多。 还是我遗漏了什么?

我不是这些领域的专家,但我会试试看是否能帮上忙。请尝试下列操作之一:

$sth = mysql_query("SELECT id, name FROM users union SELECT id, name FROM (SELECT id, title as name from another_table) as T2"); 


    $json = array(); 

        while($row = mysql_fetch_assoc($sth)) { 
            $json['name'] = $row['name']; 
            $json['id'] = $row['id']; 
        }

    $json['name'] = 'A new name';
    $json['id'] = '444';
    $data[] = $json; 

    print json_encode($data);

我从来没有做过PHP,所以我在做假设。我也从未使用过MySql,所以有更多的假设。

方案2:

$sth = mysql_query("SELECT id, name FROM users"); 


    $json = array(); 

        while($row = mysql_fetch_assoc($sth)) { 
            $json['name'] = $row['name']; 
            $json['id'] = $row['id']; 
        }

$sth = mysql_query("SELECT id, title from another_table"); 


        while($row = mysql_fetch_assoc($sth)) { 
            $json['name'] = $row['title']; 
            $json['id'] = $row['id']; 
        }

    $json['name'] = 'A new name';
    $json['id'] = '444';
    $data[] = $json; 

    print json_encode($data);