[英]How to make a dropdown with data from a mysql table?
我想通過從mysql表中獲取值來創建一個下拉區域。 這段代碼似乎不起作用; 它打印出來的是沒有內容的下拉框,我怎樣才能讓它工作? 或者這樣做是否有替代程序?
<?
$connection = mysql_connect("localhost", "root", "");
mysql_select_db("test", $connection);
$query = "SELECT full_name FROM test";
$names = mysql_query($query);
function dropDown($content, $result)
{
while($row = mysql_fetch_row($result))
{
$name = $row[0];
$content .= "<option value='$name'>$name</option>";
}
}
$content=<<<EOT
<html>
<body>
<select name="name">
EOT;
dropDown($content, $names)
$content.=<<<EOT
</select>
</body>
</html>
EOT;
echo $content;
?>
return
字符串。 PHP是不是C,你使用了 ,只是因為他們有時是很方便的參數。
function dropDown($result, $fieldName)
{
$content = '';
while($row = mysql_fetch_row($result)) {
$name = $row[0];
$content .= "<option value='$name'>$name</option>";
}
return '<select name="'.$fieldName.'">'.$content.'</select>';
}
$content = '<html><body>';
$content .= dropDown($names, 'name');
<?php echo $form->dropDownList($model,'courseprefer', array(
'(*value in database table*'=>'displaying value',
)); ?>
你可以手動添加它們只是確保數組中的第一個值在數據庫表中.. :)
避免錯誤!
here database details
mysql_connect('hostname', 'username', 'password');
mysql_select_db('database-name');
$sql = "SELECT username FROM userregistraton";
$result = mysql_query($sql);
echo "<select name='username'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['username'] ."'>" . $row['username'] ."</option>";}
echo "</select>";
here username is the column of my table(userregistration)
it works perfectly
or simply the code is
<?
$sql = "SELECT * FROM userregistraton ";
$result = mysql_query($sql); ?>
<select name="username" id="username">
<option value="">Select user</option>
<?php
while ($row = mysql_fetch_array($result))
{ ?>
<option value="<?php echo $row['uid']?>" <?php if($row['uid']==$_POST["username"]) echo "selected"; ?> >
<?php echo $row['username'];?></option>
<?php
} ?>
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.