[英]php update multiple select option values in the database
我有一个这样的数据库
CREATE TABLE IF NOT EXISTS `ia_pages` (
`pages_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`page_name` varchar(255) NOT NULL,
`search_type` varchar(120) NOT NULL,
`search_block_position` varchar(255) NOT NULL,
PRIMARY KEY (`alphabet_search_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=4 ;
INSERT INTO `ia_pages` (`pages_id`, `page_name`, `search_type`, `search_block_position`) VALUES
(1, 'home_page', 'product_search', 'right_column'),
(2, 'product_page', 'category_search', 'right_column'),
(3, 'category_page', 'product_search', 'right_column');
和这样的html形式
<table id="admin-settings">
<tbody>
<tr>
<th>Page Name</th>
<th>Search Type</th>
<th>Show Search in Block</th>
</tr>
<tr>
<td>Home Page</td>
<td>
<select id="search_type" name="search_type">
<option value="product_search">Product Search</option>
<option value="category_search">Category Search</option>
</select>
</td>
<td>
<select id="show_block" name="show_block">
<option value="left_column">Left Column</option>
<option value="right_column">Right Column</option>
</select>
</td>
</tr>
<tr>
<td>Product page</td>
<td>
<select id="search_type" name="search_type">
<option value="product_search">Product Search</option>
<option value="category_search">Category Search</option>
</select>
</td>
<td>
<select id="show_block" name="show_block">
<option value="left_column">Left Column</option>
<option value="right_column">Right Column</option>
</select>
</td>
</tr>
<tr>
<td>Category page</td>
<td>
<select id="search_type" name="search_type">
<option value="product_search">Product Search</option>
<option value="category_search">Category Search</option>
</select>
</td>
<td>
<select id="show_block" name="show_block">
<option value="left_column">Left Column</option>
<option value="right_column">Right Column</option>
</select>
</td>
</tr>
<tr>
<td style="text-align:center;" colspan="4">
<input type="submit" name="submit" value="save" class="button">
</td>
</tr>
</tbody>
</table>
现在,当我在窗体中选择任何值并单击时进行任何更改时,它将使所有行的值相同。 我的更新查询是这样的
$host = 'localhost';
$username = 'root';
$username = 'root';
$dbname = 'ia_pages';
$con=mysqli_connect($host,$username,$username);
mysqli_select_db($con,$dbname) or die ("no database");
if (mysqli_connect_errno($con)) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$search_type = $_POST['search_type'];
$show_block = $_POST['show_block'];
if(isset($_POST['submit'])) {
$update_query = "UPDATE `pages` SET `search_type` = '$search_type',`search_block_position` = '$show_block'";
$query_execute = mysqli_query($con, $update_query);
if($query_execute) {
echo "Data has been updated";
} else {
echo "data has not been updated";
}
}
即使只有选项已更改,它也会更改所有值。 因此,数据库中的所有值都已转换为相同的值。 那么如何解决呢? 我想更新在前端已更改的数据库中的值。
您需要where
更新语句中添加where
子句。 没有它,所有行都会更新。
更改:
<td>Home Page</td>
至:
<td>
Home Page
<input type="hidden" name="page_name" value="Home Page" />
</td>
将类似的更改应用于其他页面名称类型。
然后在update
查询中添加一个where
子句,如下所示:
$update_query = "UPDATE `pages` SET "
. "`search_type` = '$search_type', "
. "`search_block_position` = '$show_block' "
. "where `page_name` = '$page_name';
更新 :
如果所有选择列表都在相同的form元素中,则在php
脚本中,由于输入元素具有相同的名称,因此需要将值读入数组。
$page_names = $_POST[ 'page_name' ];
$search_types = $_POST[ '$search_type' ];
$show_blocks = $_POST[ '$show_block' ];
然后,您的更新语句应为:
update ia_pages
set search_type =
case page_name
when '$page_names[ 0 ]' then '$search_types[ 0 ]' -- Home Page
when '$page_names[ 1 ]' then '$search_types[ 1 ]' -- Product Page
when '$page_names[ 2 ]' then '$search_types[ 2 ]' -- Category Page
else search_type
end
, show_block_position =
case page_name
when '$page_names[ 0 ]' then '$show_blocks[ 0 ]' -- Home Page
when '$page_names[ 0 ]' then '$show_blocks[ 1 ]' -- Product Page
when '$page_names[ 0 ]' then '$show_blocks[ 2 ]' -- Category Page
else show_block_position
end
该查询将更新所有相关字段。
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