[英]How to Output MySql rows in Drop Down Menu
I am having a Problem in this code : ` 我的代码有问题:
<?php echo "Choose a date";
$reservation_date = $rm->retrieveReservation();
$data_date = array();
while($row_date = mysql_fetch_assoc($reservation_date)){
array_push($data_date, $row_date);
$dateOptions = ""
. "<select class='filter'>"
. "<option value='<?php echo $row['date'] ; ?>'><?php echo $row['date'] ; ?></option>"
. "</select>";
echo $dateOptions;
} ?>
?>
I want to output Sql row values in a dropdown menu. 我想在下拉菜单中输出Sql行值。 Thanks in advance!
提前致谢!
You don't have to echo the <select>
in the while, you should do like that : 您不必在一段时间内回显
<select>
,您应该这样做:
<?php echo "Choose a date";
$reservation_date = $rm->retrieveReservation();
$data_date = array();
$dateOptions = "<select class='filter'>";
while($row = mysql_fetch_assoc($reservation_date)){
array_push($data_date, $row_date);
$dateOptions = "<option value='". $row['date']."'>". $row['date'] ."</option>";
} ?>
$dateOptions = "</select>";
echo $dateOptions;
?>
This should work: 这应该工作:
echo "<select class='filter'>";
while($row = mysql_fetch_assoc($reservation_date)){
echo "<option value='".$row['date']."'>".$row['date']."</option>";
}
echo "</select>";
You need to have your select
tag outside of the loop. 您需要将
select
标签置于循环之外。 You also had a bunch of syntax errors in your PHP. 您的PHP中也有很多语法错误。
<?php echo "Choose a date";
echo "<select class='filter'>";
$reservation_date = $rm->retrieveReservation();
$data_date = array();
while($row_date = mysql_fetch_assoc($reservation_date)){
array_push($data_date, $row_date);
$dateOptions = "<option value='{$row['date']}'>{$row['date']}</option>";
echo $dateOptions;
}
echo "</select>";
?>
may this code will work for you. 也许这段代码对您有用。
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