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    2. DateTime 类的对象无法转换为字符串

      时间:2023-10-30
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                本文介绍了DateTime 类的对象无法转换为字符串的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                问题描述

                我在名为 Film_Release

                我正在循环,我想在 datetime 中转换它们并将它们展开到另一个表中.我的第二个表有一个名为 Films_Date 的列,格式为 DATE.我收到此错误

                I am looping through and I want to convert them in datetime and roll them out into another table. My second table has a column called Films_Date, with a format of DATE. I am receiving this error

                DateTime 类的对象无法转换为字符串

                Object of class DateTime could not be converted to string

                $dateFromDB = $info['Film_Release'];
                $newDate = DateTime::createFromFormat("l dS F Y",$dateFromDB); //( http:php.net/manual/en/datetime.createfromformat.php)
                

                然后我通过插入命令将 $newdate 插入到表中.

                Then I insert $newdate into the table through an insert command.

                为什么会出现这样的错误?

                Why am I be getting such an error?

                推荐答案

                因为 $newDateDateTime 类型的对象,而不是字符串.文档是明确的:

                Because $newDate is an object of type DateTime, not a string. The documentation is explicit:

                返回根据指定格式的新 DateTime 对象格式.

                Returns new DateTime object formatted according to the specified format.

                如果您想从字符串转换为 DateTime 再转换回字符串以更改格式,请调用 DateTime::format 最后从 DateTime 中获取格式化字符串.

                If you want to convert from a string to DateTime back to string to change the format, call DateTime::format at the end to get a formatted string out of your DateTime.

                $newDate = DateTime::createFromFormat("l dS F Y", $dateFromDB);
                $newDate = $newDate->format('d/m/Y'); // for example
                

                这篇关于DateTime 类的对象无法转换为字符串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跟版网!

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