Question

  if (isset($_POST["sendComments"]) ) {
    $whichProjectToSave= mysqli_real_escape_string($db5, $_POST['whichProjectToSave']);
    $commentsToPHP= mysqli_real_escape_string($db5, $_POST['commentsToPHP']);
    $objectName= mysqli_real_escape_string($db5, $_POST['objectName']);
    $sql="INSERT INTO   objectsCommentsTable (objectsCommentsText,projectName,objectName) VALUES('$commentsToPHP','$whichProjectToSave','$objectName')";
    // $sql="UPDATE objectsCommentsTable  SET objectsCommentsText ='$commentsToPHP'  WHERE projectName= '$whichProjectToSave' AND objectName = '$objectName' ";
   mysqli_query($db5,$sql);
                                  }

如果数据存在，我必须更新。但我只知道如何使用上面的方法。实际上我只能在表格中插入数据，然后通过更改代码来更新它。但我想使用类似下面的代码。如何进行？

 IF EXISTS ( SELECT objectsCommentsText FROM objectsCommentsTable  WHERE objectName='$objectName' AND projectName = '$projectName')
    UPDATE objectsCommentsTable
    SET objectsCommentsText ='$commentsToPHP'
    WHERE projectName= '$whichProjectToSave' AND objectName = '$objectName' 
ELSE 
  INSERT INTO   objectsCommentsTable (objectsCommentsText,projectName,objectName) 
  VALUES('$commentsToPHP','$whichProjectToSave','$objectName')

Answer 1

首先，添加唯一索引：

ALTER TABLE objectsCommentsTable ADD UNIQUE project_object_unique (projectName, objectName);

这将阻止添加具有相同 projectName 和 objectName 的行，并使 ON DUPLICATE 工作。

然后，您可以修改查询以使用 ON DUPLICATE

INSERT INTO objectsCommentsTable
(objectsCommentsText, projectName, objectName) VALUES ('$commentsToPHP','$whichProjectToSave', '$objectName')
ON DUPLICATE KEY
   UPDATE objectsCommentsText='$commentsToPHP'

Answer 2

您必须使用此方法INSERT ... ON DUPLICATE KEY UPDATE Syntax

如果DB表中存在数据，如何更新

2 个答案: