将特定的php变量从循环中获取到jquery函数参数

我有一个自动建议文本框女巫生成带有函数的onclick事件的li标签但我必须将php变量传递给函数作为参数但我只想要用户点击的li的变量值现在我总是得到循环中的最后一个变量。 我希望var测试值为$ test = $ row [‘t_product’]; 用户点击的li

这是现在使用的代码：

if(isset($_POST['search_term']) == true && empty($_POST['search_term']) == false){ $search_term = $_POST['search_term']; $customerid = $_SESSION['userdata']['t_customer_id']; $conn = connect(); $sql = "SELECT * FROM Licenses WHERE customer_id = '$customerid' AND t_name LIKE '$search_term%'"; $query = sqlsrv_query($conn, $sql); while(($row = sqlsrv_fetch_array($query, SQLSRV_FETCH_ASSOC)) != false){ $test = $row['t_product']; ?>  var test = ;  <?php echo '
', $row['t_name'], '
'; } } 

我希望我能很好地解释我的问题，并希望有些减少可以帮助我:)

Answer 1

好吧，也许你应该这样做：

 while(($row = sqlsrv_fetch_array($query, SQLSRV_FETCH_ASSOC)) != false){ $test = $row['t_product']; ?> ', $row['t_name'], ''; } 

要么

 if(isset($_POST['search_term']) == true && empty($_POST['search_term']) == false){ $index = 0; $search_term = $_POST['search_term']; $customerid = $_SESSION['userdata']['t_customer_id']; $conn = connect(); $sql = "SELECT * FROM Licenses WHERE customer_id = '$customerid' AND t_name LIKE '$search_term%'"; $query = sqlsrv_query($conn, $sql); while(($row = sqlsrv_fetch_array($query, SQLSRV_FETCH_ASSOC)) != false){ $test = $row['t_product']; ?>  ', $row['t_name'], ''; } } 

编辑：删除json_encode并将其替换为引号

Answer 2

通过将$ test值作为id传递给每个li并在js中获取它来解决它

  $test = $row['t_product']; echo '
', $row['t_name'], '
' function show(id){ alert($(id).attr("id"));}