Question

Unable to get a post back from this code. Get a blank white page on return

尝试输入表单并将参数传递给查询db以获取用户ID的函数。编程为php的新手。用户名密码和数据库实际值已被删除

 <?php
function capture_post_id($author_name_2, $project_name) {


$mysqli = new mysqli('localhost','username','password','database');

if ($mysqli->connect_error) 
die('Error : ('. $mysqli->connect_error .') '. $mysqli->connect_error);

  $product_name = $mysqli->query("SELECT ID FROM 'GoInst_posts' where   post_name = $project_name AND post_author = $author_name_2")->fetch_object()- >product_name; 
  print $product_name; //output value

  Global $row;
  $row = mysqli_fetch_row($product_name);

  mysqli_free_result($product_name);

  $mysqli->close();

  }


  if (isset($_POST['val1']) && isset($_POST['val2'])) {
 //$result = addNumbers(intval($_POST['val1']), intval($_POST['val2']));
 $result = capture_post_id(intval($_POST['val1'], intval($_POST['val2'])));

   }

  ?>
 <html>
 <body>

 <?php if (isset($result)) { ?>
    <h1> Result: <?php echo $result ?></h1>
 <?php } ?>

 <form action="" method="post">
 author_name_2: <input type="text" name="val1"><br>
 project_name: <input type="text" name="val2"><br>
 <input type="submit">

 </body>
 </html>

无法从mysql1 select语句返回返回

0 个答案: