如何在php中正确编写它?json
如何在php中回显json?服务器向我发送json如何在php中正确编写它?json,php,json,Php,Json,如何在php中回显json?服务器向我发送json $i=$_POST["id"]; if($i==1){ // here i must echo this { "item": { "html": [{ "description": "some text", "n": "1" }], "table": { "1": { "line": [{
$i=$_POST["id"];
if($i==1){
// here i must echo this
{
"item":
{
"html":
[{
"description": "some text",
"n": "1"
}],
"table": {
"1": {
"line": [{
"number": "",
"value": ""
}
]
}
},
"videos": [],
"urlext": [],
"imgs": [ {
"size": {
"root": "xxx xxx"
}
}]
}
}
}
else {
// here i must echo another one
{
"item":
{
"html":
[{
"description": "some text",
"n": "1"
}],
"table": {
"1": {
"line": [{
"number": "",
"value": ""
}
]
}
},
"videos": [],
"urlext": [],
"imgs": [ {
"size": {
"root": "xxx xxx"
}
}]
}
}
}
}
只需构建一个需要作为json发送的数据的关联数组,并使用
json\u encode()
呈现json
$data = array( 'name' => 'foo',
'some' => 'thing'
);
echo json_encode($data);
echo json_encode($yourContent);。你怎么没有在谷歌上找到这一点,或者甚至没有在这个网站上搜索到这一点呢???因此,这些天来,人们不断地提出一些问题,表明他们没有进行任何研究。。至少是PHP问题