PHP注意：未定义的偏移量：1，带有日期字段的字符串[重复]

Question

This question already has an answer here:

• “Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP 28 answers

I have got this issue:

Notice: Undefined offset: 1 in every row of $news_date[1].

Here are my code:

$news_date=explode("-",$press_date);
    if($news_date[1]==1){
    $news_date[1]='JAN';
    }else if($news_date[1]==2){
    $news_date[1]='FEB';
    } //and so on..
$press_date=$news_date[1].', '.$news_date[0];

I stored date data with news_date field.

Please kindly suggestion How can I should do.

Thank you very much in advance.

*** Editable**** After ask via Stack now I have used these code to solve my problem. It is just using 3 lines. Thank you Adrian Preuss for your suggestion.

    $press_date=$rs['news_date']; // store data from DB
    $new_formatDate=date("M d, Y",strtotime($press_date));  // Set Date Month Year styles that I want
    $new_formatDate = strtoupper($new_formatDate);  // Make it to ALL CAP

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Answer 1

You wan't to work with Date formats- Your problem is, you split the Date simply with explode - This way is not correct, because your splitted data has following format:

[0] "2014"
[1] "07"
[2] "21"

You check the splitted string (after splitting, each entry is also an string to) with integers:

if("07" == 1) {

The second problem is, you don't validate, if the data is set. The next problem is: You want to display only another date format - Please use PHP-based functions for formatting. Here are a sample for you:

<?php
$press_date = '2014-07-21';

print date('d. M Y', strtotime($press_date));

Output: 21. Jul 2014

<?php
$press_date = '2014-07-21';

print date('l, d. F Y', strtotime($press_date));

Output: Monday, 21. July 2014

Informations about the output format can be found at http://php.net/manual/de/function.date.php.