Php捕获致命错误 [英] Php catchable fatal error

问题描述

将$ diff插入我的数据库表时出现此错误

i have this error when inserting $diff into my database table

time_in(time)

但是即时通讯有这个错误

but im having this error

Catchable fatal error: Object of class DateInterval could not be converted to string

错误在我的查询中



我尝试过：

the error is in my query

What I have tried:

$conn = mysqli_connect(server,username,password)
        or die (mysqli_error($conn));
mysqli_select_db($conn,database)
            or die (mysqli_error($conn));

if ($now->format("H:i") > "22:00") {
    $deadline = DateTime::createFromFormat("H:i", "22:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
} else if ($now->format("H:i") > "19:00") {
    $deadline = DateTime::createFromFormat("H:i", "19:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
} else if ($now->format("H:i") > "16:00") {
    $deadline = DateTime::createFromFormat("H:i", "16:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
} else if ($now->format("H:i") > "13:00") {
    $deadline = DateTime::createFromFormat("H:i", "13:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
} else if ($now->format("H:i") > "10:00") {
    $deadline = DateTime::createFromFormat("H:i", "10:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
} else if ($now->format("H:i") > "07:00") {
    $deadline = DateTime::createFromFormat("H:i", "07:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
}

$query = "INSERT INTO `time_in`(`late`,`date_in`)
                VALUES ('$diff',NOW())";
                if (!$query) {
                printf("Error: %s\n", mysqli_error($conn));
                exit();
                } 
        if($query = $conn ->query($query)){
            echo "<br>successful";
        } else {
            echo "not successful";
        }

推荐答案

diff到我的数据库表

diff into my database table

time_in(time)

但我有这个错误

Catchable fatal error: Object of class DateInterval could not be converted to string

错误在我的查询中



我尝试过：

the error is in my query

What I have tried:

conn = mysqli_connect（服务器，用户名，密码）
或die（mysqli_error（

conn = mysqli_connect(server,username,password) or die (mysqli_error(

conn））;
mysqli_select_db（

conn)); mysqli_select_db(

这篇关于Php捕获致命错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！