我有一个JSON数组
{
"people":[
{
"id": "8080",
"content": "foo"
},
{
"id": "8097",
"content": "bar"
}
]
}
我如何搜索8097并获取内容?
答案 0 :(得分:24)
json_decode
功能可以帮助您:
$str = '{
"people":[
{
"id": "8080",
"content": "foo"
},
{
"id": "8097",
"content": "bar"
}
]
}';
$json = json_decode($str);
foreach($json->people as $item)
{
if($item->id == "8097")
{
echo $item->content;
}
}
答案 1 :(得分:17)
json_decode()
它和其他任何数组或StdClass对象一样对待
$arr = json_decode('{
"people":[
{
"id": "8080",
"content": "foo"
},
{
"id": "8097",
"content": "bar"
}
]
}',true);
$results = array_filter($arr['people'], function($people) {
return $people['id'] == 8097;
});
var_dump($results);
/*
array(1) {
[1]=>
array(2) {
["id"]=>
string(4) "8097"
["content"]=>
string(3) "bar"
}
}
*/
答案 2 :(得分:4)
如果您拥有相当少量的“人物”对象,那么之前的答案将适合您。鉴于您的示例具有8000范围内的ID,我怀疑每个ID都可能不理想。所以这里有另一种方法,可以在找到合适的人之前检查更少的人(只要人们按照ID的顺序):
//start with JSON stored as a string in $jsonStr variable
// pull sorted array from JSON
$sortedArray = json_decode($jsonStr, true);
$target = 8097; //this can be changed to any other ID you need to find
$targetPerson = findContentByIndex($sortedArray, $target, 0, count($sortedArray));
if ($targetPerson == -1) //no match was found
echo "No Match Found";
function findContentByIndex($sortedArray, $target, $low, $high) {
//this is basically a binary search
if ($high < low) return -1; //match not found
$mid = $low + (($high-$low) / 2)
if ($sortedArray[$mid]['id'] > $target)
//search the first half of the remaining objects
return findContentByIndex($sortedArray, $target, $low, $mid - 1);
else if ($sortedArray[$mid]['id'] < $target)
//search the second half of the remaining objects
return findContentByIndex($sortedArray, $target, $mid + 1, $high);
else
//match found! return it!
return $sortedArray[$mid];
}