这段代码将form value直接传递到另一个页面
<form action="table2.php" method="post">
Date : (yyyy-mm-dd)<br>
<select name="date1">
<?php
while ($row = mysqli_fetch_array($query,MYSQLI_ASSOC)){
echo "<option value='"date1'">" . $row['Date'] . "</option>";
}
?>
</select>
<br><br>
<p>Sampai</P>
Date : (yyyy-mm-dd)<br>
<select name="date2">
<?php
while ($row = mysqli_fetch_array($query,MYSQLI_ASSOC)){
echo "<option value='"date2'">" . $row['Date'] . "</option>";
}
?>
</select>
<br><br>
第一个日期工作正常，但第二个日期不会显示数据库的值。有人能帮帮我吗?由于
运行第一个mysqli_fetch_array()后，资源将为空。试试-
Date : (yyyy-mm-dd)<br>
<select name="date1">
<?php
$query1 = $query;
while ($row = mysqli_fetch_array($query1,MYSQLI_ASSOC)){
echo "<option value='"date1'">" . $row['Date'] . "</option>";
}
?>
</select>
<br><br>
<p>Sampai</P>
Date : (yyyy-mm-dd)<br>
<select name="date2">
<?php
while ($row = mysqli_fetch_array($query,MYSQLI_ASSOC)){
echo "<option value='"date2'">" . $row['Date'] . "</option>";
}
?>
</select>
<form action="table2.php" method="post">
Date : (yyyy-mm-dd)<br>
<select name="date1">
<?php
//your sql query here............
while ($row = mysqli_fetch_array($query,MYSQLI_ASSOC)){
echo "<option value='"date1'">" . $row['Date'] . "</option>";
}
?>
</select>
<br><br>
<p>Sampai</P>
Date : (yyyy-mm-dd)<br>
<select name="date2">
<?php
//again your SQL query here.........
while ($row = mysqli_fetch_array($query,MYSQLI_ASSOC)){
echo "<option value='"date2'">" . $row['Date'] . "</option>";
}
?>
</select>
<br><br>
也许您可以尝试使用JOIN语句检索date1和date2并将其循环到下拉列表中，而不是单独执行
或
select a.date1, b.date2 from table1 as, table2 as b(如果两个表之间有任何连接，可以使用ID，但这里有条件)