【发布时间】:2020-07-09 07:24:15
【问题描述】:
我有一个表。我想将输入类型日期更新为表数据。我想将其传递给 php 表单 middle.php。但似乎传递了空值,输出似乎是 1970-01-01这意味着数据没有通过。
<form method="" action="intermediate.php">
<table id="updatetable">
</table>
</form>
//Jquery added here
<script>
var today = new Date();
var dd = String(today.getDate()).padStart(2, '0');
var mm = String(today.getMonth() + 1).padStart(2, '0'); //January is 0!
var yyyy = today.getFullYear();
today = yyyy + '-' + mm + '-' + dd;
var i;
for(i=0;i<5;i++){
updatetabledata +=
'<td><input type="date" id="vazhipadudate"'+i+'"
name="vazhipadudate' + i + ' " min="' + today + '"/>
</td>';
}
$("#updatetable").empty().append(updatetabledata);
</script>
这里是intermediate.php的php代码sn-p
for($i=0;$i<5;$i++){
$date='vazhipadudate'.$i;
if(!empty( $_POST["$date"] ))
{/*echo "\n Date is ..".$_POST["$date"];*/
$time = strtotime($_POST["$date"]);
//$storecart[$i]['date']= $_POST["$date"];$storecart[$i]['date']
$time = strtotime($_POST["$date"]);
echo "The time now is.....".$time;
$storecart[$i]['date'] = date('Y-m-d', $time);
echo "Selected Date is..........".$storecart[$i]['date'] ;
}
}
输出是如果我注释掉 if(!empty( $_POST["$date"] ))
The time now is.....Selected Date is..........1970-01-01
The time now is.....Selected Date is..........1970-01-01
The time now is.....Selected Date is..........1970-01-01
The time now is.....Selected Date is..........1970-01-01
The time now is.....Selected Date is..........1970-01-01
【问题讨论】:
-
我不明白,您将输入命名为
vazhipadudate,然后您期望$_POST["$date"](注意键名中的 $),而且它似乎并不是您所拥有的全部? -
见@konrados $date='vazhipadudate'.$i
-
你输入的名字末尾有空格!还有为什么不
name="vazhipadudate[]"?