如何为jquery自动完成制作json代码？ [关闭]

问题描述：

I need an json for jquery catcomplete and I would like you to help me...

I have this code

$( ".autocomplete" ).autocomplete({
     source: "files/news.php?listar=json"
});

And I would like "files/news.php?listar=json" to display something like this:

[ 
  { label: "title1", category: "category1" },
  { label: "title2", category: "category2" } 
]

Could you please help me?

@edit:

I tried this, but it didn't work:

if ($_GET['listar'] == 'json')
{
    echo '['."
";  
    $query = mysql_query("SELECT * FROM noticias WHERE deletada='0' ORDER BY id DESC");
    $a = 0;
    while ($noticia = mysql_fetch_array($query))
    {
        echo ' { label: "'.$noticia['titulo'].'", category: "'.ucwords(Categoria($noticia['categoria'])).'" }';
        $a++;

        if ($a < mysql_num_rows($query))
        {
            echo ",
";
        }
    }
    echo "
".']';

    exit;
}

我需要一个json用于 jquery catcomplete 我想帮助你... p>

我有这段代码 p>

  $（“。autocomplete”）。selfocomplete（{
 source：“files / news.php？listar = json”
}）; 
  code>  pre> 
 
  我想“files / news.php？listar = json”显示如下内容： p> 
 
 
  [
 {label：“title1”，category：“category1”}  ，
 {标签：“标题2”，类别：“category2”} 
] 
  code>  pre> 
 
 您能帮助我吗？ p> 
 
  
 @edit： p> 
 
 
我尝试了这个，但它不起作用： p> 
 
 
  if（$ _GET ['listar'  ] =='json'）
 {
 echo'['。“
”;  
 $ query = mysql_query（“SELECT * FROM noticias WHERE deletada ='0'ORDER BY ID DESC”）; 
 $ a = 0; 
 while（$ noticia = mysql_fetch_array（$ query））
 {
  echo'{label：“'。$ noticia ['titulo']。'”，category：“'。。Copyright（Categoria（$ noticia ['categoria']））。''}'; 
 $ a ++; 
  
 if（$ a＆lt; mysql_num_rows（$ query））
 {
 echo“，
”; 
} 
} 
 echo“
”。']'; 
 
 exit;  
} 
  code>  pre> 
  div>

答

You don't need writing own json, because it's already function for json encode(): http://php.net/manual/en/function.json-encode.php

if ($_GET['listar'] == 'json')
{
    $query = mysql_query("SELECT * FROM noticias WHERE deletada='0' ORDER BY id DESC");
    while ($noticia = mysql_fetch_array($query))
    {
    $results[] = array(
        'label' => $noticia['titulo'],
        'category' => $noticia['categoria']
    );
    }
}

$json = json_encode($results);
header('Content-type: application/json');
echo $json;
exit;

答

Question is bit confusing. You are expecting JSON in PHP. In PHP, you can use build in function json_encode to convert any array into JSON and return it as response.

Your tags suggest you need it in JS. In JS, try json2.js

如何为jquery自动完成制作json代码？ [关闭]

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