大家好，我有一个问题，我正在从 abc.php 调用一个 ajax 搜索脚本，MySQL 查询工作正常，但是当我通过 ajax 搜索时，它给出了所有结果
这是 MySQL 查询:
$query = "select * from tableName where column1 like '%pattern%' or column2 like '%pattern' or column_n like '%pattern'";
这是我的 Ajax 代码
function showHint(str) {
if (str.length==0) {
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else { // code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (this.readyState==4 && this.status==200) {
document.getElementById("txtHint").innerHTML=this.responseText;
}
}
xmlhttp.open("GET","abc.php?q="+encodeURIComponent(str),true);
xmlhttp.send();
}
这是我的 abc.php
<?php
include "databaseconnect.php";
$id = trim($_GET['q']);
$res = $con->prepare('select count(*) from tablename where column = 1 AND (column = ? OR column like ? OR column like ? OR column like ? OR column like ? OR column like ? OR column like ? OR column like ?)');
$res->execute(array("$id","%$id%","%$id%","%$id%","%$id%","%$id%","%$id%","%$id%"));
if($res->fetchColumn() > 0) {
$query = $con->prepare('select * from tablename where column = 1 AND (column = ? OR column like ? OR column like ? OR column like ? OR column like ? OR column like ? OR column like ? OR column like ?)');
$query->execute(array("$id","%$id%","%$id%","%$id%","%$id%","%$id%","%$id%","%$id%"));
?>
<?php
while($row = $query->fetch(PDO::FETCH_ASSOC)){
print_r($row);
}
}
?>
这是我的 HTML 代码
<form onsubmit="return false;">
<input type="text" autocomplete="off" autofocus="autofocus" onkeyup="showHint(this.value)" class="form-control input-lg gap-top" placeholder="Search for...">
</form>
<div class="container-fluid">
<div class="table-responsive">
<p><span id="txtHint"></span></p>
最佳答案
好的，我要写下稍微不同的代码:
在MySQL中:
CREATE TABLE `test`.`words` (
`wordid` INT NOT NULL AUTO_INCREMENT,
`word_one` VARCHAR(45) NULL,
`word_tow` VARCHAR(45) NULL,
PRIMARY KEY (`wordid`));
INSERT INTO `test`.`words` (`word_one`, `word_tow`) VALUES ('a', 'd');
INSERT INTO `test`.`words` (`word_one`, `word_tow`) VALUES ('ab', 'de');
INSERT INTO `test`.`words` (`word_one`, `word_tow`) VALUES ('abc', 'def');
在 HTML 中:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>StackOverFlow</title>
<script
src="https://code.jquery.com/jquery-3.4.1.min.js"
integrity="sha256-CSXorXvZcTkaix6Yvo6HppcZGetbYMGWSFlBw8HfCJo="
crossorigin="anonymous">
</script>
<script type="text/javascript" src="app.js"></script>
</head>
<body>
<form onsubmit="return false;">
<input type="text" autocomplete="off" autofocus="autofocus" class="form-control input-lg gap-top" placeholder="Search for...">
</form>
<div class="container-fluid">
<div class="table-responsive">
<p><span id="txtHint"></span></p>
</body>
</html>在 JavaScript(app.js) 中:
$(document).ready( function(){
$('.form-control.input-lg.gap-top').keyup(function() {
let fd = new FormData();
let word = $(this).val();
fd.append('my_word', word);
$.ajax({
url: "abc.php",
type: "POST",
data: fd,
processData: false,
contentType: false,
complete: function (results) {
try {
let str = JSON.parse(results.responseText);
console.log(str)
} catch (e) {
console.error(e);
}
},
});
}) ;
});在 PHP(abc.php) 中:
<?php
$host = '127.0.0.1';
$dbname = 'test';
$username = 'kmoshe';
$password = 'password';
$charset = 'utf8';
$collate = 'utf8_unicode_ci';
$pdo = new PDO("mysql:host=$host;dbname=$dbname;charset=$charset", $username, $password,
array(
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
PDO::ATTR_PERSISTENT => false,
PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC,
PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES $charset COLLATE $collate"
)
);
$word_to_find = $_REQUEST['my_word'];
$stmt = $pdo->prepare("SELECT * FROM words WHERE word_one like :w");
$stmt->execute(['w' => "%$word_to_find%"]);
$array_of_words = [];
while($row = $stmt->fetch()) {
$array_of_words[] = $row['word_one'];
}
echo json_encode($array_of_words);关于php - 模式搜索的 Ajax 回调问题，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/47089414/