Question

希望你好......

我有一个使用php制作的API，从SQL中获取并将其转换为JSON，工作正常。我唯一的问题是，我无法操纵这个PHP来获取我想要的JSON。解决方案我认为只是在我的api.php的$ outp中的位置..我在SQL中使用id作为主要我可以用它作为索引来显示像我预期的JSON吗？

感谢你，任何努力都会得到奖励

这是输出JSON：

{
talents: [
{
id: "1",
tag: "001",
name: "Jasmina",
images: "assets/images/Jasmina3.jpg",
images02: "assets/images/Jasmina4.jpg",
images03: "assets/images/Jasmina7.jpg",
skills: "Usher",
skills02: "Modeling",
indexing: "usher"
},
{
id: "2",
tag: "002",
name: "Bruna",
images: "assets/images/BrunaD17.jpg",
images02: "assets/images/BrunaD18.jpg",
images03: "assets/images/BrunaD10.jpg",
skills: "Usher",
skills02: "Modeling",
indexing: "usher"
}
]
}

以下是我预期的结果：

{
talents: [
1: {
tag: "001",
name: "Jasmina",
images: "assets/images/Jasmina3.jpg",
images02: "assets/images/Jasmina4.jpg",
images03: "assets/images/Jasmina7.jpg",
skills: "Usher",
skills02: "Modeling",
indexing: "usher"
},

2:{
tag: "002",
name: "Bruna",
images: "assets/images/BrunaD17.jpg",
images02: "assets/images/BrunaD18.jpg",
images03: "assets/images/BrunaD10.jpg",
skills: "Usher",
skills02: "Modeling",
indexing: "usher"
}
]
}

看一看我的API，请帮我玩那个$ outp。得到它。

<?php
//header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");

$conn = new mysqli("localhost", "root", "", "application");

$result = $conn->query("SELECT * FROM talents");

$outp = "";
while($rs = $result->fetch_array(MYSQLI_ASSOC)) {
    if ($outp != "") {$outp .= ",";}
    $outp .= '{"id":"'  . $rs["id"] . '",';
    $outp .= '"tag":"'  . $rs["tag"] . '",';
    $outp .= '"name":"'   . $rs["name"]        . '",';
    $outp .= '"images":"'. $rs["images"]     . '",';
    $outp .= '"images02":"'. $rs["images02"]     . '",';
    $outp .= '"images03":"'. $rs["images03"]     . '",';
    $outp .= '"skills":"'. $rs["skills"]     . '",';
    $outp .= '"skills02":"'. $rs["skills02"]     . '",';
    $outp .= '"indexing":"'. $rs["indexing"] . '"}';
}
$outp ='{"talents":['.$outp.']}';
$conn->close();

echo($outp);
?>

Answer 1

不要试图自己构建json，而是使用native startActivityForResult()。它解决了JSON的所有边缘情况。

您需要准备的只是一个关联数组，您可以逐个字段指定（就像我在示例中所做的那样，建议使用这种方式，因为您指定要发送给用户的确切字段，它会阻止敏感数据公开）或只是$outp[$rs["id"]] = $rs;

<?php
header("Content-Type: application/json; charset=UTF-8");

$conn = new mysqli("localhost", "root", "", "application");

$result = $conn->query("SELECT * FROM talents");

$outp = array();
while ($rs = $result->fetch_array(MYSQLI_ASSOC)) {
    $outp[$rs["id"]] = array(
        "tag" => $rs["tag"],
        "name" => $rs["name"],
        "images" => $rs["images"],
        "images02" => $rs["images02"],
        "images03" => $rs["images02"],
        "skills" => $rs["skills"],
        "skills02" => $rs["skills02"],
        "indexing" => $rs["indexing"]
    );
}
$outp["talents"] = array($outp);
$conn->close();

echo (json_encode($outp));
?>

使用AngularJS处理JSON数据

1 个答案: