我在PHP中使用json_decode解码一个JSON对象，该对象可能具有1d数组或2d数组作为值：
{"Cell":{"@column":"ZjE6dW5pdmVyc2l0eQ==","@timestamp":"1425598820484","$":"MC44MDc2NDEwNDg0MjI5MjMy"}}
或者
{"Cell":[{"@column":"ZjE6YQ==","@timestamp":"1425599309809","$":"MC4wNTYzMzgwMjgxNjkwMTQwODY="},{"@column":"ZjE6YW5k","@timestamp":"1425599309809","$":"MC4wNTYzMzgwMjgxNjkwMTQwODY="},{"@column":"ZjE6Y2F0Y2hlcw==","@timestamp":"1425599309809","$":"MC4wNDIyNTM1MjExMjY3NjA1Ng=="},{"@column":"ZjE6aQ==","@timestamp":"1425599309809","$":"MC4wOTg1OTE1NDkyOTU3NzQ2NA=="},{"@column":"ZjE6dGhhdA==","@timestamp":"1425599309809","$":"MC4xNjkwMTQwODQ1MDcwNDIyNQ=="}]}
我正在使用$ Cell = $ json [“ Cell”]访问元素。我面临的问题是第二种情况下效果很好，我得到了一个数组数组，而第一种情况应该是单个元素数组，但被解释为3个元素数组。
您不能只检查元素是否具有密钥吗？像这样：
if (isset($json['Cell']['@column']) {
// do stuff with single-element
} else {
// it is a collection
}
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