Question

我希望从我的json文件中提取一个变量。我的目标是在php中使用这个变量，以便能够以mysql查询为例。

所以我的json文件在这里：pages/getRank-classic.php?idstart=0

[
    {"rank":1,"id":"111","site_name":"test1","site_vip":"No","boost":"0","site_banner":"test.png","site_banner_wallrank":"","site_pointstotaux":"5044","site_motclef1":"Pvp\/Fac","site_motclef2":"Skyblock","site_motclef3":"Cr\u00e9atif","site_motclef4":"Hunger-Games","site_motclef5":"SKywars\/Mini-Gam","site_presentationvideo":"3TGjebmNOfs"},
    {"rank":2,"id":"222","site_name":"test2","site_vip":"No","boost":"0","site_banner":"test.jpg","site_banner_wallrank":"","site_pointstotaux":"4114","site_motclef1":"hunger","site_motclef2":"games","site_motclef3":"pvp","site_motclef4":"survival","site_motclef5":null,"site_presentationvideo":"3TGjebmNOfs"}
]

我试图在包含中使用它：

<script type="text/javascript" >
$.ajax({
    type: 'POST',
    dataType: 'json',
    url: '/pages/getRank-classic.php?idstart=0',
    data: data,
    cache: false,
    success: function(test) {
        alert(test.id);
        alert(test.rank);
    }
});
</script>

那我该怎么做呢？

Answer 1

由于它是json数组，因此您还需要指定索引。

test[0].id

您也可以使用这种方式遍历对象，

    var data = [{
    "rank": 1,
    "id": "111",
    "site_name": "test1",
    "site_vip": "No",
    "boost": "0",
    "site_banner": "test.png",
    "site_banner_wallrank": "",
    "site_pointstotaux": "5044",
    "site_motclef1": "Pvp/Fac",
    "site_motclef2": "Skyblock",
    "site_motclef3": "Cr\u00e9atif",
    "site_motclef4": "Hunger-Games",
    "site_motclef5": "SKywars/Mini-Gam",
    "site_presentationvideo": "3TGjebmNOfs"
}, {
    "rank": 2,
    "id": "222",
    "site_name": "test2",
    "site_vip": "No",
    "boost": "0",
    "site_banner": "test.jpg",
    "site_banner_wallrank": "",
    "site_pointstotaux": "4114",
    "site_motclef1": "hunger",
    "site_motclef2": "games",
    "site_motclef3": "pvp",
    "site_motclef4": "survival",
    "site_motclef5": null,
    "site_presentationvideo": "3TGjebmNOfs"
    }];

    $(data).each(function () {
        alert(this.id);
    });

您可以使用ajax获取json，然后在成功事件中，将此代码添加为，

$.ajax({
type: 'POST',
dataType: 'json',
url: '/pages/getRank-classic.php?idstart=0',
data: data,
cache: false,
success: function(test) {
    $(test).each(function () {
        alert(this.id);
    });
}
});

Answer 2

我终于使用了php那样的＆gt;

<?php

$jsonlink = '/pages/getRank-classic.php?idstart=0';
$json = file_get_contents($jsonlink);

$link = mysql_connect('localhost', 'database', 'password');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}

$db_selected = mysql_select_db('database', $link);

$result = json_decode($json);
foreach($result as $key => $value) {
    if($value) {

    mysql_query("UPDATE `database`.`sites` SET `site_rank` = '$value->rank' WHERE `sites`.`site_id` = '$value->id'");
    }
    mysql_close;
}

?>

如何从json中使用jquery在php中提取变量？

2 个答案: