我可以在頁面上顯示記錄,點擊搜索按鈕後搜索也可以正常工作。現在我的問題是如何顯示按鍵上的記錄。如何使用ajax或php按鍵顯示記錄?
例如,如果有任何用戶在搜索文本字段上鍵入記錄將搜索並顯示在屏幕上的任何內容,則在同一頁上有100條記錄。
你能幫我嗎?
HTML
<script>
$(function() {
$('#valueToSearch').on('keyup',function(){
//Your ajax call will go Here
$.ajax({
url:senddata.php, // separate file for search
data : {
q : $('#valueToSearch').val().trim()
},
method:'POST',
dataType:'json',
success:function(data){
$('#fetch_record').html(data);
},
error:function(data){
alert("something has gone wrong");
}
});
});
});
</script>
<form action="" method="post">
<input type="text" name="valueToSearch" placeholder="Value To Search"><br><br>
<input type="submit" name="search" value="search"><br><br>
<table>
<tr>
<th>Id</th>
<th>First Name</th>
<th>Last Name</th>
<th>Email</th>
</tr>
<!-- populate table from mysql database -->
<?php while($row = mysqli_fetch_array($search_result)):?>
<tr>
<td><?php echo $row['id'];?></td>
<td><?php echo $row['firstname'];?></td>
<td><?php echo $row['lastname'];?></td>
<td><?php echo $row['email'];?></td>
</tr>
<?php endwhile;?>
</table>
</form>
senddata.php
if(isset($_POST['q']) && $_POST['q'] !=''){
$valueToSearch = $_POST['q'];
// your sql query for Searching result
$query = "SELECT * FROM `test` WHERE CONCAT(`id`, `firstname`, `lastname`, `email`) LIKE '%".$valueToSearch."%'";
$result = $conn->query($query);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$id=$row["id"];
$fn=$row["firstname"];
$ln=$row["lastname"];
$email=$row["email"];
}
}
return json_encode($id, $fn, $ln, $email);
}
https://www.upwork.com/hiring/development/creating-autocomplete-functionality-with-php -and-mysql/ – Exprator
我更新了我的代碼。任何人都可以幫助我嗎? –