如何删除parsererror：SyntaxError：位于0的JSON中的意外标记<

我试图使用PHP从MySQL数据库中获取数据并将其作为JSON传递。 当我尝试显示响应时，它会显示错误parsererror：SyntaxError：在位置0的JSON中出现意外的令牌。有人可以帮忙。 下面是我的jQuery和PHP代码：

jQuery的：

$(document).ready(function() { $("#display").change(function() { var type = document.getElementById('display').value; alert(type); $.ajax( { //create an ajax request to load_page.php type: "GET", url: "DBOperations.php", data : "type=" +type, dataType: "json", //expect text to be returned success: function(response) { //$("#response").html(response); //alert(response.); $.each(response, function(index, element) { $('#response').html($( { text: element.name })); }); }, error: function(jqXHR, textStatus, errorThrown) { alert('error: ' + textStatus + ': ' + errorThrown); } }); }); }); 

PHP：

 try { $dsn = 'mysql:host=localhost;dbname=practice_db'; //your host and database name here. $username = 'root'; $password = ''; //Connect to database $conn = new PDO($dsn, $username, $password); $query = "SELECT * FROM client WHERE client_type = :client_type"; //Prepare and Execute Query $stmt = $conn->prepare($query); $stmt->bindParam(':client_type', $type); $stmt->execute(); //echo 'Here: ' .$stmt; //$rows = $stmt->fetch(); $rows = $stmt->fetchAll(); foreach ($rows as $row) { echo "ClientID: ".$row['client_id'] . " "; echo "Name: ".$row['client_name'] . " "; echo "Title: ".$row['client_title'] . " "; echo "Client Type: ".$row['client_type'] . "
"; } //Display associative array echo '
'; print_r($rows) .'
'; header('Content-type: application/json'); json_encode($rows); print_r(json_encode($rows)); } catch (PDOException $ex) { echo "There was a problem executing the Query: " . $ex->getMessage(); } 
 此外，如果我尝试使用alert（）检查我得到的响应是什么，它会显示： [object HTMLDivElement] 


	







 
	
Answer 1
		      	
 使用pre标签删除回声，这些回声会污染您的输出，结果不再是有效的JSON。 


Answer 2
		      	
 你应该将json类型的字符串从php传递给jQuery，在传递的字符串中显示一些html标签，删除foreach循环并只回显json_encode($rows); 
 你的代码应该是这样的： 
 try { $dsn = 'mysql:host=localhost;dbname=practice_db'; //your host and database name here. $username = 'root'; $password = ''; //Connect to database $conn = new PDO($dsn, $username, $password); $query = "SELECT * FROM client WHERE client_type = :client_type"; //Prepare and Execute Query $stmt = $conn->prepare($query); $stmt->bindParam(':client_type', $type); $stmt->execute(); $rows = $stmt->fetchAll(); echo json_encode($rows); } catch (PDOException $ex) { echo "There was a problem executing the Query: " . $ex->getMessage(); } 
 注意：php输出在返回时不应该有任何空格或空行，例如，如果php文件如下，jQuery json解析器出错： 
 <--- here empty line --->  <--- here empty line ---> 


Answer 3
		      	



  $(document).ready(function() { $("#display").change(function() { var type = document.getElementById('display').value; $.ajax( { type: "GET", url: "check1.php", data : {"type":type}, success: function(response) { var aa=''; var data = JSON.parse(response); data.forEach(function(d){ aa+= d+'
'; }); $('#response').html(aa); }, error: function(jqXHR, textStatus, errorThrown) { alert('error: ' + textStatus + ': ' + errorThrown); } }); }); }); prepare($query); $stmt->bindParam(':client_type', $type); $stmt->execute(); $rows = $stmt->fetchAll(); $ars=null; foreach ($rows as $row) { $ars[] = $row['client_name']; } print_r(json_encode($ars)); } catch (PDOException $ex) { echo "There was a problem executing the Query: " . $ex->getMessage(); } ?> 




Answer 4
		      	
 从php页面获取json时只在页面上打印json。 
 尝试删除此部分 
 echo'
'; print_r($rows).'
';
 
 希望这可以帮助。