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[英]selected value get from db into dropdown select box option using php mysql error
[英]selected value get from db into Data list option using php mysql
我需要从 db 中获取选定的值到 datalist 框中。告诉我怎么做。 这是代码。
<input list="Rank_Name" class="form-control" required>
<datalist id="Rank_Name">
<?php
$sel_cus = "select Rank_Name from ranks where Rank_Status=1";
$res_cus = mysqli_query($connection, $sel_cus);
while ($row = mysqli_fetch_array($res_cus)) {
?>
<option value="<?php echo $row['Rank_Name'];?>"></option>
<?php
}
?>
</datalist>
如果我理解正确,您还需要在下拉列表中选择具有其他值的值。 你可以通过这样做来实现这一点
<?php
$select1="select Rank_Name from ranks where Rank_Status=1";
$q=mysqli_query($select1) or die($select1);
$row=mysqli_fetch_array($q); //here you are getting name of person whose rank is 1
?>
<datalist id="Rank_Name">
<?php
$s="select * from ranks ";
$q=mysqli_query($s) or die($s);
while($r=mysqli_fetch_array($q))
{ ?>
<option value="<?php echo $r['Rank_Name']; ?>"<?php if($row['Rank_Name']==$r['Rank_Name']) echo 'selected="selected"'; ?>>
<?php echo $r['Rank_Name']; ?>
</option>
<?php } ?>
</datalist>
在上面的代码中,这一行<?php if($row['Rank_Name']==$r['Rank_Name']) echo 'selected="selected"'; ?>
<?php if($row['Rank_Name']==$r['Rank_Name']) echo 'selected="selected"'; ?>
检查第一次查询的值是否相同,如果相同则该选项将被自动selected
<input list="Rank_Name" class="form-control" required>
<datalist id="Rank_Name">
<?php
$sel_cus = "select Rank_Name from ranks where Rank_Status=1";
$res_cus = mysqli_query($connection, $sel_cus);
while ($row = mysqli_fetch_array($res_cus)) {
echo "<option value=".$row['Rank_Name']."></option>";
}
?>
</datalist>
试试这个代码。 我在 while 循环中使用 echo <option>
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