Question

我使用get方法从mysql数据库获取值我从url传递survey_id和question_id

如下 http://myserver.com/emrapp/surveyAnswersScreenOne.php?survey_id=1,question_id=1

但是它给出了错误

我的php代码在下面给出了

     $query = mysql_query("SELECT * from survey_Answers where survey_Id='".$survey_id."' AND question_Id='".$question_id"' ");


     $rows = array();

     while($row = mysql_fetch_assoc($query)) {

     $rows[] = $row;
     }

     echo json_encode($rows);

Answer 1

此行有错误，显示：

$question_id"' "

应该是

$question_id . "'"

Answer 2

你应该用＆amp;分隔获取变量不，在网址中。

Answer 3

那是因为你应该在url参数之间使用&而不是,

http://myserver.com/emrapp/surveyAnswersScreenOne.php?survey_id=1&question_id=1

$survey_id = mysql_real_escape_string($_GET['survey_id']);
$question_id = mysql_real_escape_string($_GET['question_id']);

Answer 4

GET参数通常用&而不是,分隔。所以你的链接看起来应该更像这样：

http://myserver.com/emrapp/surveyAnswersScreenOne.php?survey_id=1&question_id=1

另外，请注意GET变量不会自动转换为PHP变量。你需要从$_GET数组中取出它们：

$survey_id = $_GET['survey_id']

Answer 5

你没有传递在url中分隔的值逗号。你有使用＆amp;在网址中

http://myserver.com/emrapp/surveyAnswersScreenOne.php?survey_id=1&question_id=1

Answer 6

 $query = mysql_query("SELECT * from survey_Answers where survey_Id='".$survey_id."' AND question_Id='".$question_id"'");

你忘记了在$question_id - ＆gt;之后连接字符串的重点。这应该可以解决您的问题：

$query = mysql_query("SELECT * from survey_Answers where survey_Id='".$survey_id."' AND question_Id='".$question_id."'");

无论如何，还要考虑清理你的网址 - ＆gt; http://xkcd.com/327/

Answer 7

 $query = mysql_query("SELECT * from survey_Answers where survey_Id='".$survey_id."' AND question_Id='".$question_id."' ");

缺少一个点！

Answer 8

网址和查询错误

http://myserver.com/emrapp/surveyAnswersScreenOne.php?survey_id=1&question_id=1

AND

$question_id . "' "

使用get方法从mysql数据库获取值

8 个答案: