PHP/MySQLi错误 - 不会插入或更新

-2

我已经尝试了几个版本的下面的脚本，并没有拿出一个可行的解决方案。我基本上尝试一个try/catch来运行不同的MySQLi查询。执行时，我没有收到任何错误，但没有任何内容正在插入或更新到我的数据库中。PHP/MySQLi错误 - 不会插入或更新

$monitor_insert = "INSERT INTO monitors (serial,room,deployed,asset) VALUES ('$monitor','5','$timestamp','$asset')" or die("Error " . mysqli_error($conn)); 
     if (!mysqli_query($conn, $monitor_insert)) { 
      $monitor_update = "UPDATE monitors SET serial = '$monitor', room = '5', deployed = '$timestamp', asset = '$asset')" or die("Error " . mysqli_error($conn)); 
      if (!mysqli_query($conn, $monitor_update)) { 
       echo "<p>Nothing Worked!!!</p>"; 
      } 
      else { 
       $audit_update = "INSERT INTO audit (code,info) VALUES ('15',Monitor $monitor has been updated to deployed by $uid')"; 
       mysqli_query($conn, $audit_update); 
      } 
     } 
     else { 
      $audit_update = "INSERT INTO audit (code,info) VALUES ('17',Monitor $monitor has been added by manual scan by $uid')"; 
      mysqli_query($conn, $audit_update); 

      $audit_update = "INSERT INTO audit (code,info) VALUES ('15',Monitor $monitor has been updated to deployed by $uid')"; 
      mysqli_query($conn, $audit_update); 
     }

第1-11行是不工作的行。

来源

2017-04-25 Joseph Monarch

添加一些调试语句。检查并打印_each_ query的响应，以便知道您要关闭的路径。我假设'mysqli_query（）'返回false，而你忽略它。 –

阅读您的查询，看起来好像您的报价不正确，您缺少字符串上的开引号。 '（'15'，'Monitor $ monitor已更新为由$ uid部署'）' –

谢谢@MattClark。失踪的报价是我倒下的秋天。我想我只需要第二组眼睛。 –

工作脚本：

$monitor_insert = "INSERT INTO monitors (serial,room,deployed,asset) VALUES ('$monitor','5','$timestamp','$asset')" or die("Error " . mysqli_error($conn)); 
     if (!mysqli_query($conn, $monitor_insert)) { 
      $monitor_update = "UPDATE monitors SET serial = '$monitor', room = '5', deployed = '$timestamp', asset = '$asset')" or die("Error " . mysqli_error($conn)); 
      if (!mysqli_query($conn, $monitor_update)) { 
       echo "<p>Nothing Worked!!!</p>"; 
      } 
      else { 
       $audit_update = "INSERT INTO audit (code,info) VALUES ('15','Monitor $monitor has been updated to deployed by $uid')"; 
       mysqli_query($conn, $audit_update); 
      } 
     } 
     else { 
      $audit_update = "INSERT INTO audit (code,info) VALUES ('17','Monitor $monitor has been added by manual scan by $uid')"; 
      mysqli_query($conn, $audit_update); 

      $audit_update = "INSERT INTO audit (code,info) VALUES ('15','Monitor $monitor has been updated to deployed by $uid')"; 
      mysqli_query($conn, $audit_update); 
     }

来源

2017-04-25 22:23:44

PHP/MySQLi错误 - 不会插入或更新

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