Php 从将字符串分隔为键的JSON获取值
我有一个JSON,我想从中获取值。但是,键值是典型的键。它们不是一个词Php 从将字符串分隔为键的JSON获取值,php,arrays,json,Php,Arrays,Json,我有一个JSON,我想从中获取值。但是,键值是典型的键。它们不是一个词 [ { "quality-of-service": "Good" }, { "quality-of-staff": "Great" }, { "quality-of-communication": "Excellent" }, { "value-for-money": "Excellent" } ]
[
{
"quality-of-service": "Good"
},
{
"quality-of-staff": "Great"
},
{
"quality-of-communication": "Excellent"
},
{
"value-for-money": "Excellent"
}
]
如何获取键的值。解码
$array=json\u decode($json,true)
后,第一个值将是$array[0]['quality-of-service']]
,这不是一个好方法。您可以循环:
foreach($array as $values) {
echo key($values) . " is " . current($values);
}
或者可以展平阵列:
$array = array_merge(...$array);
然后使用$array['quality-of-service']
或循环它:
foreach($array as $key => $value) {
echo "$key is $value";
}
与任何其他数组@abracadver相同,那么如何单独获取键
人员素质的值呢?