我已经被这个问题困扰了几天了。 我要开发的系统逻辑在这里
这是图像的代码
<?php
session_start();
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Search</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<!-- <link rel="stylesheet" type="text/css" href="style.css"/> -->
</head>
<body>
<form action="process/searchprocess.php" method="GET">
<table width="100%">
<tr>
<th>
Client Ic<br><input type="text" name="client_name" />
</th>
<th>
Client Ic<br><input type="text" name="client_ic" />
</th>
<th>
Client Address <br><input type="text" name="client_add" />
</th>
</tr>
</table>
<table width="100%">
<tr>
<th>
<br><input type="submit" value="Search" align="center" />
</th>
</tr>
</table>
</form>
</body>
</html>
这是我的流程代码
<?php
session_start();
mysql_connect("localhost", "root", "") or die("Error connecting to database: ".mysql_error());
mysql_select_db("waveevo") or die(mysql_error());
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Search results</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<!-- <link rel="stylesheet" type="text/css" href="style.css"/> -->
<style>
table tr:nth-child(even) {
background-color: #eee;
}
table tr:nth-child(odd) {
background-color:#fff;
}
table th {
background-color: black;
color: white;
}
</style>
</head>
<body>
<?
php
$query = $_GET['client_name'];
$query2 = $_GET['client_ic'];
$query3 = $_GET['client_add'];
if ($query == null && $query2 == null && $query3 == null)
{
echo "Please at least insert one the value";
}
else
{
$query = htmlspecialchars($query);
$query2 = htmlspecialchars($query2);
$query3 = htmlspecialchars($query3);
$query = mysql_real_escape_string($query);
$query2 = mysql_real_escape_string($query2);
$query3 = mysql_real_escape_string($query3);
$raw_results = mysql_query("SELECT * FROM client WHERE ('client_name' LIKE '%".$query."%') OR ('client_ic' LIKE '%".$query2."%') OR ('client_add_1' && ' ' && 'client_add_2' && ' ' && 'client_add_3' && ' ' && 'client_add_4' LIKE '%".$query3."%')") or die(mysql_error());;
if(mysql_num_rows($raw_results) > null){ // if one or more rows are returned do following
while($results = mysql_fetch_array($raw_results)){
?>
<table width="100%">
<tr>
<th>ID</th>
<th>Name</th>
<th>IC</th>
<th>Mobile</th>
<th>Address</th>
<th>Marital Status</th>
<th>Race</th>
<th>Asset Type</th>
<th>Bank</th>
<th>Amount</th>
<th>Nationality</th>
<th>Limit</th>
</tr>
<tr>
<td><?php echo $results["client_id"]; ?></td>
<td><?php echo $results["client_name"]; ?></td>
<td><?php echo $results["client_ic"]; ?></td>
<td><br><?php echo $results["client_mobile_1"]."<br>".$results["client_mobile_2"]."<br>".$results["client_mobile_3"]; ?></td>
<td><?php echo $results["client_add_1"]."<br>".$results["client_add_2"]."<br>".$results["client_add_3"]."<br>".$results["client_city"]."<br>".$results["client_postcode"]; ?></td>
<td><?php echo $results["client_marital_status_id"]; ?></td>
<td><?php echo $results["client_race_id"]; ?></td>
<td><?php echo $results["client_asset_type_id"]; ?></td>
<td><?php echo $results["client_bank_id"]; ?></td>
<td><?php echo $results["amount"]; ?></td>
<td><?php echo $results["client_nationality_id"]; ?></td>
<td><?php echo $results["client_limit"]; ?></td>
</tr>
</table>
<?php
}
}
else{ // if there is no matching rows do following
echo "No results";
}
}
?>
</body>
所以我尝试输入客户端IC的值，例如1234，数据库中没有数据与我刚才输入的值匹配，但结果仍然显示有，我可以知道为什么吗，因为我已经没有解决这个问题的方法
最佳答案
那是因为当你使用通配符时 LIKE '%".$query."%' 如果你的变量 $query 为空，你就会得到全部，因为你与一切都像“%%”。
您需要更改这句话:
$raw_results = mysql_query("SELECT * FROM client WHERE ('client_name' LIKE '%".$query."%') OR ('client_ic' LIKE '%".$query2."%') OR ('client_add_1' && ' ' && 'client_add_2' && ' ' && 'client_add_3' && ' ' && 'client_add_4' LIKE '%".$query3."%')") or die(mysql_error());;
有这样的事情
$sql_query="SELECT * FROM client WHERE ";
$other=false;
if ($query != null and $query!="") {
$sql_query=$sql_query."('client_name' LIKE '%".$query."%')";
$other=true;
}
if ($query2 != null and $query2!="") {
if ($other) {
$sql_query=$sql_query." OR ";
}
$sql_query=$sql_query."('client_ic' LIKE '%".$query2."%')";
$other=true;
}
if ($query3 != null and $query3!="") {
if ($other) {
$sql_query=$sql_query." OR ";
}
$sql_query=$sql_query."('client_add_1' && ' ' && 'client_add_2' && ' ' && 'client_add_3' && ' ' && 'client_add_4' LIKE '%".$query3."%')";
$other=true;
}
if ($other) {
$raw_results = mysql_query($sql_query) or die(mysql_error());
}
关于javascript - 使用 php 和 mysql 表单上的多个搜索字段，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44237228/