我已经为此工作了几个小时，但毫无结果。
我正在尝试使用 PHP 和 mysql 为我的网站创建高级搜索功能。
高级搜索中只有复选框。
复选框如下所示:
<form action="search.php" method="POST" id="myForm"/>
<table width="100%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td> </td>
<td><input type="checkbox" name="keyword[]" value="small"/></td>
<td>Small</td>
</tr>
<tr>
<td> </td>
<td><input type="checkbox" name="keyword[]" value="large"/></td>
<td>Large</td>
</tr>
<tr>
<td> </td>
<td><input type="checkbox" name="keyword[]" value="xlarge"/></td>
<td>X-Large</td>
</tr>
<tr>
<td> </td>
<td><input type="checkbox" name="keyword[]" value="xxlarge"/></td>
<td>XX-Large</td>
</tr>
<tr>
<td> </td>
<td> </td>
<td> </td>
</tr>
<tr>
<td> </td>
<td><input type="checkbox" name="keyword[]" value="red"/></td>
<td>Red</td>
</tr>
<tr>
<td> </td>
<td><input type="checkbox" name="keyword[]" value="black"/></td>
<td>Black</td>
</tr>
<tr>
<td> </td>
<td><input type="checkbox" name="keyword[]" value="white"/></td>
<td>White</td>
</tr>
</table>
<input type="submit" value="submit" />
</form>
我的 MYSQL 数据库如下所示:
产品表:
id product_name price
5 shirt 20
6 shoe 70
我有一个属性表，用于存储产品的颜色和尺寸。
该表如下所示:
idc id product_name colors sizes
1 5 shirt red
2 5 shirt back
3 5 shirt white
4 5 shirt small
5 5 shirt medium
6 5 shirt Large
7 6 shoe brown
8 6 shoe black
9 6 shoe 5
10 6 shoe 6
11 6 shoe 7
在我的 search.php 页面中，我有以下代码:
include "config/connect.php";
$searchList = "";
foreach($_POST['keyword'] as $c){
$sql ="SELECT *
FROM `yt`
INNER JOIN `ATTRIBUTES` ON yt.id= ATTRIBUTES.id
WHERE (
ATTRIBUTES.size LIKE '%".$c."%' OR
ATTRIBUTES.color LIKE '%".$c."%'
)";
$query = mysqli_query($db_conx, $sql);
}
//var_dump($query);
$productCount = mysqli_num_rows($query);
$i=0; // count the output amount
if ($productCount > 0) {
while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){
$id = $row["id"];
$product_name = $row["product_name"];
$searchList .= ''.$product_name.'';
}
}
上面的代码没有返回任何内容，并且我不断收到此错误:
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given
这意味着我的sql查询后出了问题!
所以我做了var_dump($query);，这给了我bool(fales)。
没什么帮助，因为这与第一个错误的含义相同!
有人可以帮我解决这个问题吗？
如有任何建议和帮助，我们将不胜感激。
编辑:
我已将代码更改为:
<?php
error_reporting(-1);
ini_set('display_errors', 'On');
include "config/connect.php";
$searchList = "";
$clause = " WHERE ";//Initial clause
$sql="SELECT *
FROM `yt`
INNER JOIN `ATTRIBUTES` ON yt.id=ATTRIBUTES.id";//Query stub
if(isset($_POST['submit'])){
if(isset($_POST['keyword'])){
foreach($_POST['keyword'] as $c){
if(!empty($c)){
$sql .= $clause."`".$c."` LIKE '%{$c}%'";
$clause = " OR ";//Change to OR after 1st WHERE
} $query = mysqli_query($db_conx, $sql);
}
}
//echo $sql;//Remove after testing
}
//var_dump($query);
$productCount = mysqli_num_rows($query);
$i=0; // count the output amount
if ($productCount > 0) {
while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){
$id = $row["id"];
$product_name = $row["product_name"];
$searchList .= ''.$product_name.'';
}
}
?>
<?php echo $searchList; ?>
但是上面的代码给了我以下错误:
Notice: Undefined variable: query in on line 23
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given on line 23
我的方向正确吗？
编辑:
此代码从 MYSQL 数据库返回数据，但信息错误!
$clause = " WHERE ";//Initial clause
$sql="SELECT *
FROM `yt`
INNER JOIN `ATTRIBUTES` ON yt.id=ATTRIBUTES.id";//Query stub
if(isset($_POST['submit'])){
if(isset($_POST['keyword'])){
foreach($_POST['keyword'] as $c){
if(!empty($c)){
$sql .= $clause."`sizes` LIKE '%{$c}%'";
$clause = " OR ";//Change to OR after 1st WHERE
}
}
// Run query outside of foreach loop so it only runs one time.
$query = mysqli_query($db_conx, $sql);
// Check that the query ran fine.
if (!$query) {
print "ERROR: " . mysqli_error($db_conx);
} else {
// Use $query inside this section to make sure $query exists.
$productCount = mysqli_num_rows($query);
$i=0; // count the output amount
if ($productCount > 0) {
while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){
$id = $row["id"];
$product_name = $row["product_name"];
$searchList .= ''.$product_name.'';
}
}
}
}
}
最佳答案
尝试将其更改为这样，看看您会得到什么:
<?php
error_reporting(-1);
ini_set('display_errors', 'On');
include "config/connect.php";
$searchList = "";
$clause = " WHERE ";//Initial clause
$sql="SELECT *
FROM `yt`
INNER JOIN `ATTRIBUTES` ON yt.id=ATTRIBUTES.id";//Query stub
if(isset($_POST['submit'])){
if(isset($_POST['keyword'])){
foreach($_POST['keyword'] as $c){
if(!empty($c)){
##NOPE##$sql .= $clause."`".$c."` LIKE '%{$c}%'";
$sql .= $clause . " (ATTRIBUTES.sizes LIKE '%$c%' OR ATTRIBUTES.colors LIKE '%$c%')";
$clause = " OR ";//Change to OR after 1st WHERE
}
}
print "SQL Query: $sql<br />"; //<-- Debug SQl syntax.
// Run query outside of foreach loop so it only runs one time.
$query = mysqli_query($db_conx, $sql);
var_dump($query); //<-- Debug query results.
// Check that the query ran fine.
if (!$query) {
print "ERROR: " . mysqli_error($db_conx);
} else {
// Use $query inside this section to make sure $query exists.
$productCount = mysqli_num_rows($query);
$i=0; // count the output amount
if ($productCount > 0) {
while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){
$id = $row["id"];
$product_name = $row["product_name"];
$searchList .= ''.$product_name.'';
}
}
}
}
}
?>
<?php echo $searchList; ?>
已编辑:修复了错误的列名称。 也许是这样的？
关于php - 使用复选框搜索 mysql？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/28721675/