PHP回显乱序

Question

我用 PHP 编写了一个脚本，它从 MySQL 数据库中获取值并将这些值输出为 JSON。但是，我遇到了问题。这是代码:

    <?php 
    mysql_connect("localhost", "root", "");
    mysql_select_db("stitchdex");
    $query = mysql_query("SELECT image, price, description FROM {$_GET['table']}");
    $jsonString = "{\"images\":[";
    $numrows = mysql_num_rows($query);
    $i=0;
    while ( $row = mysql_fetch_array($query) ) {
        $jsonString .= "{\"image\":\"{$row['image']}\", \"price\":\"{$row['price']}\", \"description\":\"{$row['description']}\"}";
        if ( ($numrows-1) == $i ) echo "";
        else echo ",";
        $i++;
    }
    $jsonString .= "]}";
    echo $jsonString;
?>

但它输出:

,,,,,,,,,,,,,,,,,{"images":[{"image":"priceguide1.jpg", "price":"", "description":"Diagram of Options"}{"image":"2ndamendfront.jpg", "price":"$109.00", "description":"Total of 3 laser etchings"}{"image":"DSC_0120.jpg", "price":"$99.00", "description":"Designer Kydex 1 Laser Etching"}{"image":"DSC_0121.jpg", "price":"", "description":"Lazer Etching on Back"}{"image":"gilbert.jpg", "price":"$109.00", "description":"Designer Kydex 2 Lazer Etchings"}{"image":"glock23-carbonfiber-red1.jpg", "price":"$79.00", "description":""}{"image":"irish1.jpg", "price":"$129.00", "description":"Designer Kydex Two Tone Leather & 3 Laser Etchings Two on front One on Back"}{"image":"irish2.jpg", "price":"$129.00", "description":"Designer Kydex Two Tone Leather & 3 Laser Etchings Two on front One on Back"}{"image":"klover-glock-2.jpg", "price":"$129.00", "description":"Designer Kydex Two Tone Leather & 3 Laser Etchings Two on front One on Back"}{"image":"priceguide1.jpg", "price":"", "description":"Diagram of Options"}{"image":"2ndamendfront.jpg", "price":"$109.00", "description":"Total of 3 laser etchings"}{"image":"DSC_0120.jpg", "price":"$99.00", "description":"Designer Kydex 1 Laser Etching"}{"image":"DSC_0121.jpg", "price":"", "description":"Lazer Etching on Back"}{"image":"gilbert.jpg", "price":"$109.00", "description":"Designer Kydex 2 Lazer Etchings"}{"image":"glock23-carbonfiber-red1.jpg", "price":"$79.00", "description":""}{"image":"irish1.jpg", "price":"$129.00", "description":"Designer Kydex Two Tone Leather & 3 Laser Etchings Two on front One on Back"}{"image":"irish2.jpg", "price":"$129.00", "description":"Designer Kydex Two Tone Leather & 3 Laser Etchings Two on front One on Back"}{"image":"klover-glock-2.jpg", "price":"$129.00", "description":"Designer Kydex Two Tone Leather & 3 Laser Etchings Two on front One on Back"}]}

如果逗号分隔“图像”数组中的对象，这将是有效的 JSON，但出于某种原因，脚本首先输出所有逗号。这是为什么？

Answer 1

我建议不要手动构建 json 字符串，为此使用 json_encode()，这就是它存在的原因。

$data = array();
while ( $row = mysql_fetch_assoc($query) ) {
    $data[] = $row;
}

echo json_encode(array('images' => $data));

旁注:另外，使用改进的扩展，即 mysqli 或 PDO，并利用准备好的语句。您当前的代码直接在查询中使用 $_GET 变量，容易受到 sql 注入(inject)的攻击。绑定(bind)它们以生成更安全的查询。