我的所有 html 都工作正常，但当我尝试自动完成字段时，我的 php 代码似乎有问题
search.php
<!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>
<link href="css/style.css" rel="stylesheet" type="text/css">
<SCRIPT LANGUAGE="JavaScript" src="js/jquery.js"></SCRIPT>
<SCRIPT LANGUAGE="JavaScript" src="js/script.js"></SCRIPT>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<div class="main">
<div class=""><a href="http://www.scriptime.blogspot.in">scriptime</a></span></div>
<div id="holder">
Enter Keyword : <input type="text" id="keyword" tabindex="0"><img src="images/loading.gif" id="loading">
</div>
<div id="ajax_response"></div>
</div>
</body>
</html>
这是我的 php 代码
名称.php
<?php
include("Connections/myphp.php");
$keyword = $_POST['data'];
$sql = "select username from ".$users." where ".$username." like '".$keyword."%' limit 0,20";
//$sql = "select username from ".$users."";
$result = mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($result))
{
echo '<ul class="list">';
while($row = mysql_fetch_array($result))
{
$str = strtolower($row['username']);
$start = strpos($str,$keyword);
$end = similar_text($str,$keyword);
$last = substr($str,$end,strlen($str));
$first = substr($str,$start,$end);
$final = '<span class="bold">'.$first.'</span>'.$last;
echo '<li><a href=\'javascript:void(0);\'>'.$final.'</a></li>';
}
echo "</ul>";
}
else
echo 0;
?>
ajax代码
/*
cc:scriptime.blogspot.in
edited by :midhun.pottmmal
*/
$(document).ready(function(){
$(document).click(function(){
$("#ajax_response").fadeOut('slow');
});
$("#keyword").focus();
var offset = $("#keyword").offset();
var width = $("#keyword").width()-2;
$("#ajax_response").css("left",offset.left);
$("#ajax_response").css("width",width);
$("#keyword").keyup(function(event){
//alert(event.keyCode);
var keyword = $("#keyword").val();
if(keyword.length)
{
if(event.keyCode != 40 && event.keyCode != 38 && event.keyCode != 13)
{
$("#loading").css("visibility","visible");
$.ajax({
type: "POST",
url: "names.php",
data: "data="+keyword,
success: function(msg){
if(msg != 0)
$("#ajax_response").fadeIn("slow").html(msg);
else
{
$("#ajax_response").fadeIn("slow");
$("#ajax_response").html('<div style="text-align:left;">No Matches Found</div>');
}
$("#loading").css("visibility","hidden");
}
});
}
else
{
switch (event.keyCode)
{
case 40:
{
found = 0;
$("li").each(function(){
if($(this).attr("class") == "selected")
found = 1;
});
if(found == 1)
{
var sel = $("li[class='selected']");
sel.next().addClass("selected");
sel.removeClass("selected");
}
else
$("li:first").addClass("selected");
}
break;
case 38:
{
found = 0;
$("li").each(function(){
if($(this).attr("class") == "selected")
found = 1;
});
if(found == 1)
{
var sel = $("li[class='selected']");
sel.prev().addClass("selected");
sel.removeClass("selected");
}
else
$("li:last").addClass("selected");
}
break;
case 13:
$("#ajax_response").fadeOut("slow");
$("#keyword").val($("li[class='selected'] a").text());
break;
}
}
}
else
$("#ajax_response").fadeOut("slow");
});
$("#ajax_response").mouseover(function(){
$(this).find("li a:first-child").mouseover(function () {
$(this).addClass("selected");
});
$(this).find("li a:first-child").mouseout(function () {
$(this).removeClass("selected");
});
$(this).find("li a:first-child").click(function () {
$("#keyword").val($(this).text());
$("#ajax_response").fadeOut("slow");
});
});
});
当我尝试搜索名称时，它给我一个错误:您的 SQL 语法有错误；检查与您的 MySQL 服务器版本相对应的手册，了解在第 1 行的“where like 'ben%' limit 0,20”附近使用的正确语法
最佳答案
$sql = "select username from ".$users." where ".$username." like '".$keyword."%' limit 0,20";
您错过了 $users 和 $username 它们的值是什么？
关于PHP 自动完成功能不起作用，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20165839/