将数据从mysqli拉到下拉列表php
[英]pull the data from mysqli to dropdown list php
问题描述
I have been trying to pull the data from database mysqli in the drop down menu. 
我一直在尝试从下拉菜单中的数据库mysqli中提取数据。 Here is my code for postad.html. 这是我的postad.html代码。 I want to pull categories data from categories table with a row name CatName. 我想从具有行名CatName的类别表中提取类别数据。 I also have CatId that shows ctegory id. 我也有CatId显示ctegory ID。 Thank you in advance.Please help: 预先谢谢您，请帮助：
<html>
<head>
<title>Post AD</title>
</head>
<body>
<form method="POST" action="postad.php">
<table>
<tr>
<td>AdName:</td>
<td><input type="text" name="adname"></td>
</tr>
<tr>
<td>AdCategory</td>
<td>
//dropdown list
<select name="Categories">
<?php include = connect.php
$sql = "SELECT CatName FROM categories order by CatName";
$result = $db->query($sql);
while($row = $result->fetch_array()) {
echo "<option value='". $row['CatName']."'>."</option>';
}
?>
</select>
</td>
</tr>
<tr>
<td>Contact Number</td>
<td><input type="text" name="contactnumber"></td>
</tr>
<tr>
<td>Image</td>
<td><input type="text" name="image"></td>
</tr>
<tr>
<td>Description</td>
<td><input type="text" name="description"></td>
</tr>
<tr>
<td>Expiration Date</td>
<td><input type="text" name="expirationdate"></td>
</tr>
<tr>
<td id="btn"><input type="submit" value='submit' name="submit"></td>
</tr>
</table>
</form>
</body>
</html>
3 个解决方案
解决方案1
1 2014-11-13 03:23:45
Looks like you have an error with your menu code. 菜单代码似乎有误。 Should be this 应该是这个
echo "<option value=". $row['CatName'] ."></option>';
Opposed to
echo "<option value='". $row['CatName']."'>."</option>';
The result from the table is going to be a string regardless so there is no need to worry about value=''. 该表的结果将是一个字符串，因此无需担心value =''。 The real problem was how you muffled up your quotes. 真正的问题是您如何低估报价。
EDIT: Just noticed include = connect.php should be include 'connect.php'; 编辑：刚注意到include = connect.php应该是include 'connect.php';
<?php
$mysql_host = '';
$mysql_user = '';
$mysql_pass = '';
$mysql_dbase = '';
$mysql_table = '';
$pdo = new PDO('mysql:host=' . $mysql_host . ';dbname=' . $mysql_dbase, $mysql_user, $mysql_pass);
$sth = $pdo->prepare('SELECT * FROM `' . $mysql_table . '` ORDER BY `id` CatName');
$sth->execute();
$result = $sth->fetchAll();
foreach ($result as $row) {?>
<option value=<?php echo $row['CatName']; ?>></option>;
<?}?>
解决方案2
0 2014-11-13 03:41:37
echo "<option value='". $row['CatName']."'>."</option>';
contains error and also there is no inner html for options 包含错误，也没有内部html选项
You can use this way for simplicity. 您可以使用这种方式来简化操作。
$CatName=$row["CatName"];
echo '<option value="'.$CatName.'">'.$CatName.'</option>';
解决方案3
0 2014-11-13 05:29:06
Ok. 好。 Now I have created a new php file called getcategories.php. 现在，我创建了一个名为getcategories.php的新php文件。 and this page works well. 此页面效果很好。 Here is the code for it: 这是它的代码：
$sql = "SELECT CatName FROM categories order by CatName";
$result = $db->query($sql);
$options="";
//echo "<select value='categories'>";
while($row = $result->fetch_assoc()) {
$categoryname=$row["CatName"];
// echo '<option value="'.$categoryname.'">'.$categoryname.'</option>';
// echo "</select>";
//echo "$categoryname\n";
$options .= '<option value="'.$categoryname.'">'.$categoryname.'</option>';
}
?>
How should I pull the data of options to the dropdown of html file 我应该如何将选项数据拉到html文件的下拉列表中
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