Question

我已经在数据库中创建了一个表，用户可以在其中存储有关自身的日常信息，数据已成功保存到表中，但是如果我尝试更新表中已有的数据就会出错（如果用户已经在当天输入了信息，而不是创建新行）。

  $sql = "SELECT * FROM $username WHERE day=?;";
  // Here we initialize a new statement by connecting to the database (dbh.php file)
  $stmt = mysqli_stmt_init($conn);
  if (!mysqli_stmt_prepare($stmt, $sql)) {
    // If there is an error the user is sent to the enter data page again
    header("Location: ../enterTodaysData.php?error=sqlerror");
    exit();
  }
  else { //if there are no errors...
    mysqli_stmt_bind_param($stmt, "s", $day); //binds the parameters to the statement
    mysqli_stmt_execute($stmt); //executes the statement

    $result = mysqli_stmt_get_result($stmt); //saves the result of the statement into the result variable

    if ($row = mysqli_fetch_assoc($result)) { //if the user HAS already made an entry that day
      $sql = "UPDATE $username (SET peakflow1 = $peakflow1 WHERE day=$day);";
      $sql = "UPDATE $username (SET peakflow2 = $peakflow2 WHERE day=$day);";
      $sql = "UPDATE $username (SET coughing = $coughing WHERE day=$day);";
      $sql = "UPDATE $username (SET tightChest = $tightChest WHERE day=$day);";
      $sql = "UPDATE $username (SET shortBreath = $shortBreath WHERE day=$day);";
      $sql = "UPDATE $username (SET wheezing = $wheezing WHERE day=$day);";
      $sql = "UPDATE $username (SET symptomOne = $symptomOne WHERE day=$day);";
      $sql = "UPDATE $username (SET symptomTwo = $symptomTwo WHERE day=$day);";
      $sql = "UPDATE $username (SET medication = $medication WHERE day=$day);";
      $sql = "UPDATE $username (SET mood = $mood WHERE day=$day);";
      $sql = "UPDATE $username (SET comments = $comments WHERE day=$day);";
      $sql = "UPDATE $username (SET overall = $overall WHERE day=$day);";
      header("Location: ../home.php?sql=success");
      exit();
    }
    else{ //if the user has not
      $sql = "INSERT INTO $username (day,   peakflow1,  peakflow2,  medication, mood,   coughing,   tightChest, shortBreath,    wheezing,   symptomOne, symptomTwo, overall,    comments) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?);"; //the question marks are placeholders
      $stmt = mysqli_stmt_init($conn);
      //an sql statement is prepared and the database is connected to
      if (!mysqli_stmt_prepare($stmt, $sql)) {
        // If there is an error the user is sent back to the signup page
        header("Location: ../enterTodaysdata.php?error=sqlerror");
        exit();
      }
      else {
        //binds the paramaters and data to the statement
        mysqli_stmt_bind_param($stmt, "siisiiiiiiiis", $day,    $peakflow1, $peakflow2, $medication,    $mood,  $coughing,  $tightChest,    $shortBreath,   $wheezing,  $symptomOne,    $symptomTwo,    $overall,   $comments);
        //this executes the prepared statement and send it to the database, this registers the user.
        mysqli_stmt_execute($stmt);
        //sends the user back to the signup page, with a message confirming that it was a success
        header("Location: ../home.php?sql=success");
        exit();
      }
    }
  }
}

导致该问题的代码部分从$sql = "UPDATE $username (SET peakflow1 = $peakflow1 WHERE day=$day);";行开始。除了屏幕顶部的“ sqlerror”错误消息外，屏幕上没有显示任何结果，但是表不会更新。

Answer 1

我发现更新存在一些问题。（在运行时可能还会出现其他问题，但这就是我在您发布的代码中看到的。）

在此部分：

if ($row = mysqli_fetch_assoc($result)) { //if the user HAS already made an entry that day
    $sql = "UPDATE $username (SET peakflow1 = $peakflow1 WHERE day=$day);";
    $sql = "UPDATE $username (SET peakflow2 = $peakflow2 WHERE day=$day);";
    ...

这些$sql = "UPDATE ...表达式中的每一个都将覆盖$sql变量，因此在该部分的结尾$sql仅保留最后一个查询。

SET peakflow1 ...等括号最多是不必要的，我认为它们会导致SQL语法错误。
这些SQL字符串中的任何变量都没有引号，并且其中一些包含字符串。这将导致SQL语法错误。（请参见When to use single quotes, double quotes, and backticks in MySQL。）应通过与执行插入操作相同的方式执行更新，并通过将变量绑定到预准备语句中的占位符来避免此问题。顺便说一句，您可以使用单个查询来完成所有更新，例如：
```
"UPDATE $username SET peakflow1 = ?, peakflow2 = ?, ... WHERE day = ?"
```
您没有执行该SQL。您可以将SQL字符串分配给$sql变量，然后立即exit()对其进行任何操作。

使用此处的操作，您可以通过执行“ UPSERT”来简化代码，基本上不必运行三个查询（检查是否存在，如果不存在则进行插入，如果是则进行更新），则可以运行一个将插入或更新的查询。在MySQL中，只要day列定义为唯一，就可以使用INSERT... ON DUPLICATE KEY UPDATE语法。

尝试更新表中的信息时出现错误吗？

1 个答案: