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我用jQuery數據變量發佈閱讀,但PHP是無法讀取數據變量的Json字符串化不能用PHP
JavaScript端:
var data = JSON.stringify(details);
$.ajax({
data: data,
type: 'POST',
url: '/dashboard/example.php',
});
PHP端:
$json=json_decode($_POST['data']);
foreach ($json as $value) {
...SQL update...
}
如果我這樣在PHP中聲明數據變量,它是有效的,但發佈不起作用。
$string='[{"node":{"_DT_RowIndex":1},"oldData":"2","newData":"1","newPosition":0,"oldPosition":1},{"node":{"_DT_RowIndex":2},"oldData":"3","newData":"2","newPosition":1,"oldPosition":2},{"node":{"_DT_RowIndex":3},"oldData":"4","newData":"3","newPosition":2,"oldPosition":3},{"node":{"_DT_RowIndex":0},"oldData":"1","newData":"4","newPosition":3,"oldPosition":0}]';
$json=json_decode($_POST['data']);
foreach ($json as $value) {
...
}
你能幫我嗎?
var dump'$ _POST ['data']'解碼之前,看看它是什麼? –
只需打印出$ _POST ['data']'並相應地工作。或嘗試一次'data:''「+ data +」'「,' –
我無法在PHP文件中回顯,因此var_dump無法工作 – hamozoo