如何将选定内容添加到此代码中

时间:2017-04-23 21:23:14

标签: php mysql input option

This code works.
<!doctype html>
<body>
<form  method="post"  >
<?php

$conn = new mysqli('localhost', 'root', '', 'myDB');

if ($conn->connect_error){
die("Connection failed: " . $conn->connect_error);
} 

$sql='SELECT DISTINCT counselor_name FROM counselors';
$result = $conn->query($sql);

?>

<select name="name" >
<?php
while ($row =  mysqli_fetch_assoc($result)) {
echo '<option value="'.$row['counselor_name'].'">'.$row['counselor_name'] . '</option>';
}
?>
</select>
<input type="submit" value="GO!"  name="go"/>
</form>
</body>

对于我的生活,我无法弄清楚如何将select函数合并到选项字符串中以选择值。 IE:如果我要设置$ filter =&#34; John&#34;。我想&#34; John&#34;被选中。

提前致谢

1 个答案:

答案 0 :(得分:0)

您可以使用三元运算符在一行(现有的回显线)中进行,但为了便于阅读,这就是您想要的:

if ($row['counselor_name'] === $filter)
{
  echo '<option value="'.$row['counselor_name'].'" selected="selected">'.$row['counselor_name'] . '</option>';
}
else
{
  echo '<option value="'.$row['counselor_name'].'">'.$row['counselor_name'] . '</option>';
}