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“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP
                                
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                            mysqli_fetch_assoc() expects parameter / Call to a member function bind_param() errors. How to get the actual mysql error and fix it?
                                
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                                2年前关闭。
            
                    
查询1:

$query = "SELECT myname, DATE_FORMAT(birthdate, '%m/%d/%Y') FROM `personal_info`";


在此查询中,它有效,并且给了我所需的格式(dd / mm / yyyy),但是在另一个查询中,给了我一条错误消息。

查询2(不在同一页面上):

$query = "SELECT `myname`, `private_ID`, DATE_FORMAT(`birthdate`, '%m/%d/%Y'), `adress`, `phoneno`, `email`, `id` FROM `personal_info` WHERE `private_ID` = $id LIMIT 1";


$result = mysqli_query($conn, $query);

// if id exist
// show data in inputs
if(mysqli_num_rows($result) > 0)
{
  while ($row = mysqli_fetch_array($result))
  {
    $myname= $row['myname'];
    $private_ID = $row['private_ID '];
    $birthdate= $row['birthdate'];
    $adress= $row['adress'];
    $phoneno = $row['phoneno '];
    $email= $row['email'];
    $id= $row['id'];
  }
}


(此查询2用于填充从MySQL获取的值,并将其插入HTML表单字段。)

注意:未定义的索引:第31行和第31行的---的生日是:

$birthdate= $row['birthdate'];


我的错在哪里?

最佳答案

将您的查询更改为此:

DATE_FORMAT(`birthdate`, '%m/%d/%Y') AS birthdate

关于php - “YYYY-MM-DD”到“DD-MM-YYYY”在一个查询中有效,在第二个查询中有效-不会,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/48371588/

10-14 17:41
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