I have this function in php
我在php中有这个功能
<?php function showTable(){
$url = "http://10.0.0.1/lib/api/desk/branch/";
$params = array ("action" => "list","company_key" => "1");
$result=requestURL($url,$params);
$json_a=json_decode(strip_tags($result),true);
?>
This is the json I got from the code above. I need to check the key "status". If it has a value "no", it should display an alert that status is not ok. Thanks!
这是我从上面的代码得到的json。我需要检查关键“状态”。如果它的值为“no”,则应显示状态不正常的警报。谢谢!
{
"init": [
{
"status": "ok",
"record_count": 9,
"code": "",
"message": "",
"page_count": null,
"current_page": 0
}
]
}
2 个解决方案
#1
1
Easy.
Suppose you hold the JSON in $json
.
假设你在$ json中持有JSON。
$jsonDecoded = json_decode($json, true);
if($jsonDecoded['init'][0]['status'] != 'ok') {
print "status is not ok :(";
}
#2
0
Let say that json is in $json variable
假设json在$ json变量中
$json = json_decode($json,true);
$status = $json['init'][0]['status'];
if($status=='ok'){
# do something here
}else{
# display not ok here
}
#1
1
Easy.
Suppose you hold the JSON in $json
.
假设你在$ json中持有JSON。
$jsonDecoded = json_decode($json, true);
if($jsonDecoded['init'][0]['status'] != 'ok') {
print "status is not ok :(";
}
#2
0
Let say that json is in $json variable
假设json在$ json变量中
$json = json_decode($json,true);
$status = $json['init'][0]['status'];
if($status=='ok'){
# do something here
}else{
# display not ok here
}