设置标签的值<;选择>;php中数据库的html
下面的代码设置从表department选择的标记的值设置标签的值<;选择>;php中数据库的html,php,html,database,Php,Html,Database,下面的代码设置从表department选择的标记的值 <select name='dept'> <option value="">Select a department</option> <?php $sql = "SELECT deptNo, deptName FROM Department"; $result = mysqli_query($conn, $sql); if (mysqli_num_
<select name='dept'>
<option value="">Select a department</option>
<?php
$sql = "SELECT deptNo, deptName FROM Department";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
echo "<option value=$row[deptNo]";
选择一个部门
试试这个
if ($dept == $row["deptNo"]) echo ' selected="selected"';
echo语句中的空格尝试以下操作:
<?php
$sql = "SELECT deptNo, deptName FROM Department";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$selected = ($dept == $row["deptNo"]) ? "selected" : "";
?>
<option value= "<?php echo $row['deptNo']; ?>" <?php echo $selected; ?> >
<?php echo $row['deptName']; ?>
</option>
<?php } } ?>
希望这有帮助
和平!xD
<?php
$sql = "SELECT deptNo, deptName FROM Department";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$selected = ($dept == $row["deptNo"]) ? "selected" : "";
?>
<option value= "<?php echo $row['deptNo']; ?>" <?php echo $selected; ?> >
<?php echo $row['deptName']; ?>
</option>
<?php } } ?>