Question

我开始学习PHP，因为VBA还不够。今天我请你帮忙。我有Select框，它从MySQL获取数据。我有输入，我想从第二列输入数据，同一个表将数据带到选择框，不知道我是怎么做的。

Answer 1

如果要将值从变量添加到输入字段，可以执行下一步

//$var you get from query from this your mysql table
<input type="text" name="somename" value="<?php echo $var['second_column']; ?>" id="someid">

Answer 2

您是否在询问如何将数据放入数据库中的选择框？如果是这样，你可以使用它..

<select name="country">
<option value="">Select</option>
<?php
$countriesQuery=mysqli_query($conn,"SELECT id,name FROM countries");
while($countries=mysqli_fetch_assoc($countriesQuery))
{
echo "<option value='$countries[id]'>$countries[name]</option>";
}
?>
</select>

Answer 3

<select name="country">
<option value="">Select</option>
<?php
$countriesQuery=mysqli_query($conn,"SELECT id,name FROM countries");
while($countries=mysqli_fetch_assoc($countriesQuery))
{
    echo "<option value='".$countries[id]."'>".$countries[name]."   </option>";
}
?>
</select>

Answer 4

希望这有帮助

<?php
$mysqli= new mysqli("localhost", "root", "", "db");
if ($mysqli->connect_errno) {
  printf("Connect failed: %s\n", $mysqli->connect_error);
  exit();
}
$result=$mysqli->query("select name from table");
/* name is your second field*/
?>
<select name="">
<?php

while($res=$result->fetch_array())
{?>
  <option value="<?php echo $res[0]; ?>"><?php echo $res[0]; ?></option>
 <?php
}
?>
</select>

PHP / MySQL选择框和输入

4 个答案: