Character Wise使用AJAX和PHP进行搜索?

时间:2016-08-18 15:45:57

标签: javascript php mysql sql ajax

我已经完成了从正面匹配角色的高级搜索。我希望它也应该与单词的中间和结尾匹配,或者与数据库中的任何地方匹配。我怎么能知道我的预先搜索代码在这里 PHP代码........................

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
<style>
table {
    width: 100%;
    border-collapse: collapse;
}

table, td, th {
    border: 1px solid black;
    padding: 5px;
}

th {text-align: left;}
</style>
</head>

<body>
<?php
		$servername="localhost";
		$username="root";
		$password="";
		$conn=mysql_connect($servername,$username,$password);
	if (!$conn) {
    	die('Could not connect: ' . mysqli_error($conn));
	}
	
	$in=$_GET['q'];
if(!ctype_alnum($in)){
    echo "Data Error";
    exit;
}

		mysql_select_db('firstdb');
		$sql="select name, id , age, sex from name where name like '$in%' or sex like '$in%' or age like '$in%' or id like '$in%'";
		$display=mysql_query($sql,$conn);
	echo "<table>
	<tr>
	<th>ID</th>
	<th>Name</th>
	<th>Age</th>
	<th>Gender</th>
	</tr>";
while( $row = mysql_fetch_assoc( $display ) ){
    echo "<tr>";
    echo "<td>" . $row['id'] . "</td>";
	echo "<td>" . $row['name'] . "</td>";
    echo "<td>" . $row['age'] . "</td>";
    echo "<td>" . $row['sex'] . "</td>";
    echo "</tr>";
}
echo "</table>";

?>
</body>
</html>

JS代码

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
<script>
function user(str){
    if(str.length==0)
	{
	document.getElementById("userhint").innerHTML = "";
        return;
	}	
	else {
        var xmlhttp = new XMLHttpRequest();
        xmlhttp.onreadystatechange = function() {
            if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                document.getElementById("userhint").innerHTML = xmlhttp.responseText;
            }
        }
        xmlhttp.open("GET", "advanceSearch.php?q="+str, true);
        xmlhttp.send();
    }

	
}

</script>
</head>
<body>

<form>
<input type="text" onkeyup="user(this.value);" name="username" />
</form>
<br />
<div id="userhint"><b>User info will be listed here...</b></div>
</body>
</html>

1 个答案:

答案 0 :(得分:1)

通过查询中的微小变化来完成

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$sql="select name, id , age, sex from name where name like '%$in%' or sex like '$%in%' or age like '%%$in%' or id like '%$in%'";
&#13;
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