Question

所以我的问题是标题中的错误，有我的代码：

jQuery(document).ready(function ($) {
	$("#food_search").keyup(function(event){
		var search_term =$(this).val();
$.ajax({
	type:"POST",
	url:"Mypage",
	data:{'fsearch':search_term},
	success:function(res){
		$("#food_search_result").html(res);
		console.log(res);
	},
	error: function (xhr, ajaxOptions, thrownError) {
           alert(xhr.status);
           alert(xhr.responseText);
           alert(thrownError);
       }
});
	});
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<!----------------------------------------------------------------
                              HTML
----------------------------------------------------------------->

<form method="post">
<p>Търсене на храни: <input type="text" name="fsearch" id="food_search"></p>
</form>
<!----------------------------------------------------------------
                              PHP
----------------------------------------------------------------->
<?php

$hostname = "localhost";
$username = "name";
$password = "pass";
$databaseName = "dbname";
$connect = mysqli_connect($hostname, $username, $password, $databaseName);


if(!empty($_POST["fsearch"])) {
$req = $connect->prepare('SELECT * FROM food_data_bg WHERE title LIKE "%".$fsearch."%"');

$req->execute(array(
'title'=>'%'.$_POST['fsearch'].'%'
));
if($req->rowCount()==0){
echo 'Не бяха намерени резултати!';
}
else{
while($data=$req->fetch()){
?>
<div class="search-result">
    <img src="<?php echo $data['fimage']; ?>" class="fimage"/>
    <span class="result-title"><?php echo $data['title'] ;?></span><br>
    <span class="calories-total"><?php echo $data['calories total'] ;?></span><br>
</div>
<?php
}
}
}
?>

代码用于即时搜索（google like）。$req变量必须只获取db中的信息，就像jquery通过ajax传递给php的变量search_term一样，php必须检查是否有单词或短语对应于输入字段中写的东西。

感谢！

Answer 1

$fsearch = $connect->real_escape_string($_POST['fsearch']);
$req = $connect->query("SELECT * FROM food_data_bg WHERE title LIKE %{$fsearch}%");
while($row=$req->fetch_array()){
    //output function  data: $row['title'], $row['calories total'], etc
}

mysqli docs

Answer 2

我通过添加var $ fsearch并修复我的查询调用来修复它，因为“％”。$ fsearch。“％”'不正确我写了'“。$ fsearch。”％'“正如我所说像这样添加了var $ fsearch：$fsearch = $_POST['fsearch'];我尝试过使用

if($req) {
//code if it return true (if it is working)
}else{
//code if it return false(not working)
}

现在它回归真实，这意味着查询正在运行，但给了我其他的

错误：mysqli_stmt :: execute（）需要0个参数，给定1个

我会问其他问题。谢谢你的帮助！

在非对象上调用成员函数execute（）

2 个答案: