Question

我正在尝试使用 http://loopj.com/jquery-tokeninput

我的PHP代码如下

$data = array();
$result=$this->ArticleTag->query("SELECT * FROM tbl_article_tags where name LIKE '".$_GET['q']."%'");
foreach ($result as $row){
$name ='id:'.$row['tbl_article_tags'] ['id'].',name:'.$row['tbl_article_tags']['name'];
array_push($data, $name);   
}   
echo json_encode($data);

给出了输出

["id :57, name :Editorial The Corporate","id :15, name :editorial","id :93, name :Editorial from abroad"]

我需要以下列格式输出JSON搜索结果：

[
    {"id":"856","name":"House"},
    {"id":"1035","name":"Desperate Housewives"},
    ...
]

我尝试了不同的组合，但没有奏效。请帮忙。

Answer 1

Try this:完整代码解决方案＆gt;你的问题出在$name变量..

    $data = array();
    $result=$this->ArticleTag->query("SELECT * FROM tbl_article_tags where name LIKE '".$_GET['q']."%'");
    foreach ($result as $row){
    $name = array(
          'id' => $row['tbl_article_tags']['id'],
          'name' => $row['tbl_article_tags']['name']
           );
        $data[]=$name;   
    }   
    echo json_encode($data);

Answer 2

试试此代码

string json_encode ( mixed $value [, int $options = 0 [, int $depth = 512 ]] )

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

Answer 3

$name = array('id' => $row['tbl_article_tags']['id'], 'name' => $row['tbl_article_tags']['name']);

json_encode适用于多维数组

Answer 4

您可以使用2个数组，因此值将配对。

$data = array();
foreach ($result as $row) {
    $tempArray = array(
            ['id'] = $row['tbl_article_tags'],
            ['name'] = $row['tbl_article_tags']['name']);
    array_push($data, $tempArray);   
}   
echo json_encode($data);

格式化JSON输出

4 个答案: