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我有一个包含一些字段的表格。地址就是其中之一。我使用 textarea 标签作为地址。现在的问题是，当我根据一个地址搜索某些东西时，它只适用于一行地址。我的意思是，如果我在地址中输入了换行符（段落格式），它不会显示结果。所有其他搜索查询都运行良好。这是我的代码，

地址下拉列表代码：

<input type="hidden" name="category_address" value="address"/>
<select name='criteria_address' style="width:100px;">
    <option selected="selected"></option>
    <?php
    $order = "SELECT DISTINCT address FROM lh_clients ORDER BY clientname" or die (mysql_error());
    $result = mysql_query($order);
    while ($data = mysql_fetch_array($result)) {
        echo ("<option>$data[address]</option>");
    }
    ?>
</select>

搜索查询代码：

if (isset($_POST['criteria_address']))
{
$category_address = $_POST['category_address'];
$criteria_address = $_POST['criteria_address'];
$query = "SELECT * FROM lh_clients WHERE $category_address = '$criteria_address' ORDER BY clientname";
$result   = mysql_query($query) or die(mysql_error());
$num_rows = mysql_num_rows($result);


while ($data = mysql_fetch_array($result)) { ?>
<tr>
<td style="vertical-align:top;"><? echo $data['clientname']; ?></td>
<td style="vertical-align:top;"><? echo $data['clienttype']; ?></td>
<td style="vertical-align:top;"><? echo nl2br($data['address']); ?></td>
<td style="vertical-align:top;"><? echo $data['contacts']; ?></td>
<td style="vertical-align:top;"><? echo $data['sensitivity']; ?></td>
<td style="vertical-align:top;"><? echo $data['acountmanager']; ?></td>
<td style="vertical-align:top;"><? echo $data['responsibleexecutive']; ?></td>
<td></td>
</tr>
<?php
}

}

我认为搜索查询不适用于地址中的换行符。

4

2 回答 2

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您可以尝试LIKE像这样使用运算符：

WHERE $category_address LIKE '{$criteria_address}%'

这将找到以 $criteria 地址开头的所有记录。

于 2012-07-19T12:33:23.920 回答
0

出色地...

$order = "SELECT DISTINCT address FROM lh_clients ORDER BY clientname" or die (mysql_error());
$result = mysql_query($order);
while ($data = mysql_fetch_array($result)) {
    foreach (explode("\n", $data[address]) as $line) {
        echo ("<option value='".htmlentities($line)."'>{$line}</option>");
    }
}

尝试使用 LIKE 在文本列中进行搜索

$address = trim(html_entity_decode($criteria_address));
$query = "SELECT * FROM lh_clients WHERE $category_address like '%{$address}%' or $category_address like '{$address}%' or $category_address like '%{$address}' ORDER BY clientname";
于 2012-07-19T12:34:43.893 回答