所以我的問題是錯誤到標題上,有我的代碼:調用一個成員函數的execute()非對象
jQuery(document).ready(function ($) {
\t $("#food_search").keyup(function(event){
\t \t var search_term =$(this).val();
$.ajax({
\t type:"POST",
\t url:"Mypage",
\t data:{'fsearch':search_term},
\t success:function(res){
\t \t $("#food_search_result").html(res);
\t \t console.log(res);
\t },
\t error: function (xhr, ajaxOptions, thrownError) {
alert(xhr.status);
alert(xhr.responseText);
alert(thrownError);
}
});
\t });
});
<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
}
}
}
?>
代碼用於即時搜索(谷歌喜歡)。 $req
變量必須只從db獲取信息,就像變量search_term通過jquery通過ajax傳遞給php,並且php必須檢查是否有與輸入字段中寫入的東西相對應的單詞或短語。
謝謝!
我把這段代碼添加到我的代碼中,它說'mysqli_connect'找不到 –
對不起,它是'$ connect = new mysqli($ hostname,$ username,$ password,$ databaseName);'你必須在面向對象時尚太 –