PHP echo html table error

我已经使用数据库中的数据创建了一个表来创建它我使用循环从我需要的列中获取每一条数据。现在，我在输入的HTMLonClick中调用了javascript函数。但是，我不断收到错误。该功能似乎工作，这只是实际的echo，似乎是给出了错误。它在第58行，你可以看到我正在创建一个表并插入行。PHP echo html table error

UPDATE

我的JavaScript功能正在工作。错误是当它试图设置<td>宽度和对齐。我得到的错误是：

未捕获的SyntaxError：无效的或意外的标记

<?php 
    ob_start(); 
    session_start(); 
    require_once 'dbconnect.php'; 
    if(!isset($_SESSION['user'])) { 
     header("Location: index.php"); 
     exit; 
    } 
    $deny = array("222.222.222", "333.333.333"); 
    if (in_array ($_SERVER['REMOTE_ADDR'], $deny)) { 
     header("location: http://www.google.com/"); 
     exit(); 
    } 
    $res=mysqli_query($con,"SELECT * FROM users WHERE userId=".$_SESSION['user']); 
    $userRow=mysqli_fetch_array($res); 
?> 
<!DOCTYPE html> 
<html> 
    <?php header("Access-Control-Allow-Origin: http://www.py69.esy.es"); ?> 
    <head> 
     <title>ServiceCoin</title> 
     <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
     <link rel="stylesheet" href="assets/css/bootstrap.min.css" type="text/css" /> 
     <link rel="stylesheet" href="scripts/home/index.css" /> 
    </head> 
    <body> 
     <ul> 
      <li><a href="#" class="a">ServiceCoin.com(image)</a></li> 
      <li><a href="logout.php?logout" class="a">Sign Out</a></li> 
      <li><a href="#" class="a">Contact</a></li> 
      <li><a href="#" class="a">Get Service Coins</a></li> 
      <li><a href="#" class="a">News</a></li> 
      <li><a href="settings.php" class="a">Settings</a></li> 
      <li><a href="#" class="a">Referrals</a></li> 
      <li><a href="service.php" class="a">Services</a></li> 
      <li><a href="home.php" class="a">Home</a></li> 
     </ul> 
     <br /><br /> 
     <center> 
     <h3>Welcome, <?php echo $userRow['userName']; ?>. You Currently Have <span id="services"><?php echo $userRow['userServices']; ?></span> Services</h3> 
     <p id="error"></p> 
     <button onclick="send_coins()" class="button">Send Coins</button> 
     <button onclick="create_service()" class="button">Create A Service</button> 
     <button onclick="send_coins()" class="button">My Services</button> 
     <h3>View Services</h3> 
     <span><?php 
     $users = 1; 
     $num = 1; 
     echo "<center><table width=1000><th>Buy</th><th>Service Name</th><th>Service Cost</th><th>Service Description</th><th>Service Provider Name</th>"; 
     while($users < 100 && $num < 100){ 
      $res=mysqli_query($con,"SELECT * FROM users WHERE userId=".$users); 
      $userRow=mysqli_fetch_array($res); 
      $id = 0; 
      while($id != 10){ 
       $id = $id + 1; 
       if($userRow['userService'.$id] === 'null'){ 
       }else if(!empty($userRow['userService'.$id])){ 
       echo "<tr class=services ><td name=".$num."><input type=submit onClick='buy(".$userRow['userService'.$id].",".$userRow['userServiceCost'.$id].",".$userRow['userServiceEmail'].")'></td><td width='250' align='center'>".$userRow['userService'.$id]."</td><td width='250' align='center'>".$userRow['userServiceCost'.$id]."</td><td width='250' align='center'>".$userRow['userServiceDes'.$id]."</td><td width='250' align='center'>".$userRow['userServiceName'.$id]."</td></tr>"; 

       //echo $id."Error: ".$con->error;   
       $num = $num + 1; 
       } 
      } 
      $users = $users + 1; 
     } 
     echo "All Services Loaded"; 
     echo "</table></center>"; 
     ?></span> 
      <span class="text-danger"><?php echo $msg; ?></span> 
     </center> 
    </body> 
    <script lang="text/javascript" src="scripts/home/index.js"></script>  
    <script type="text/javascript"> 


function buy(sid, scost, semail){ 
    console.log(sid,scost,semail); 
} 

</script> 
</html> 
<?php ob_end_flush(); ?>

来源

2016-11-17 Morgan Clarke

的可能的复制[如何从PHP调用JavaScript函数？（http://stackoverflow.com/questions/1045845/how-to-call-a-javascript-function-from-php ） –

看看编辑请 –

''td width ='250'align ='center'>“'把单引号放在所示位置，看看有没有帮助 – Blinkydamo

正如我在上面提到的意见，你需要使用引号的所有属性和JS调用。此外，请注意您的onClick的内容必须是有效的JS。因此，作为参数传递给函数的字符串应该用引号引起来。

echo '<tr class="services"><td name="'.$num.'"><input type="submit" onClick="buy(\''.$userRow['userService'.$id].'\',\''.$userRow['userServiceCost'.$id].'\',\''.$userRow['userServiceEmail'].'\')"></td><td width="250" align="center">'.$userRow['userService'.$id].'</td><td width="250" align="center">'.$userRow['userServiceCost'.$id].'</td><td width="250" align="center">'.$userRow['userServiceDes'.$id].'</td><td width="250" align="center">'.$userRow['userServiceName'.$id].'</td></tr>';

来源

2016-11-17 20:04:22 Steve

是的，抱歉，我注意到了，忘记了更新这里的代码。对不起，这并没有解决它 –

PHP echo html table error

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