Question

我试图从数据库中获取值并在下拉菜单中显示它们但是我在下拉菜单中什么都没有，这里有代码请任何人吗？

<select name="car" value="Select" size="1">

<?php
$sql = "SELECT fullname FROM users"; 

    $result = mysql_query($sql) or die (mysql_error()); 

    while ($row = mysql_fetch_array($result))

    {
            $name=$row['fullname']; 
            $options.="<OPTION VALUE=\"$name\">";
    }

?>

<option>
<? echo $options ?>
</option>
</select>

Answer 1

您的代码中有多处错误。试试这个：

<select name="car" value="Select" size="1">

<?php
$sql = "SELECT fullname FROM users"; 

    $result = mysql_query($sql) or die (mysql_error()); 

    while ($row = mysql_fetch_array($result))
    {
            //securing from XSS
            $name= htmlentities($row['fullname']); 

            //you had no closing tag, no name
            $options.="<OPTION VALUE=\"$name\">$name</option>";
    }
    //need semicolon, no need for tags
    echo $options;
?>

</select>

Answer 2

你的意思是

<select name="car" value="Select" size="1">
<? echo $options; ?>
</select>

$ options 也应采用以下格式：

$options .= '<option value="'.$name.'">'.$name.'</option>';

Answer 3

in your code Semicolon(;) is missing in the end of $options:

<option>
<? echo $options ?>
</option>

also remove `value="Select"` with select tag

try this:

<select name="car" size="1">

<?php
    $sql = "SELECT fullname FROM users";     
    $result = mysql_query($sql) or die (mysql_error()); 

    while ($row = mysql_fetch_array($result))

    {
            $name=$row['fullname']; 
           echo "<OPTION VALUE='$name'>$options</option>";
    }

?>
</select>

Answer 4

您错过了选项完整语法$options.="<OPTION VALUE=\"$name\">"; 应为$options.="<OPTION VALUE=\"$name\">$name</OPTION>";

编辑

<option><!--< not required as you already used in $options-->
<? echo $options ?>
</option><!--< not required as you already used in $options-->

Answer 5

将您的代码更改为此..

<select name='cars'> <?php $sql = "SELECT fullname FROM users"; 

$result = mysql_query($sql) or die (mysql_error()); 

while ($row = mysql_fetch_array($result))

{
        echo "<OPTION VALUE='".$row['fullname']."'>".$row['fullname']."</OPTION>";
}?></select>

Answer 6

如果您在表中有条目，只需从代码中删除额外选项

<select name="car" value="Select" size="1">

<?php
$sql = "SELECT fullname FROM users"; 

    $result = mysql_query($sql) or die (mysql_error()); 

    while ($row = mysql_fetch_array($result))

    {
            $name=$row['fullname']; 
            $options.="<OPTION VALUE=\"$name\">";
    }

?>
<? echo $options ?>
</option>
</select>

在<option>

之前删除<? echo $options ?>

从数据库下拉菜单没有结果

6 个答案: