选择echo echo只在工作日播放php

Question

I have the below code which displays only weekdays in php.

<?php for($i=0;$i<=4;$i++): ?>
<label ><?php echo date('l',strtotime("+$i Weekday")); ?>,</label>
<label ><?php echo date('M d', strtotime("+$i Weekday")); ?></label>
<br>   
<?php endfor; ?>

The code displays

Thursday, Sep 17 
Friday, Sep 18 
Monday, Sep 21 
Tuesday, Sep 22 
Wednesday, Sep 23 

My goal is to display the weekdays in the below format;

Today --- How can I achieve this
Tomorrow --- How can I achieve this
Monday, Sep 21
Tuesday, Sep 22
Wednesday, Sep 23

I will like to use it in a <select> html tag and display the dates as options. Any help will be appreciated.

Cheers!

Answer 1

Try this code:

<?php
$today = date('c');
$tomorrow = date('c', strtotime('+1 day'));

for($i=0;$i<=4;$i++){ ?>
<label><?php 
    if(date('c',strtotime("+$i day")) == $today){
        echo "Today";
    } elseif(date('c',strtotime("+$i days")) == $tomorrow){
        echo "Tomorrow";
    } else {
        echo date('l',strtotime("+$i day"));
        echo ", </label><label>";
        echo date('M d', strtotime("+$i day"));
    }
    ?></label>
<br>   
<?php } ?>

Output :

EDIT : Don't show Saturday and Sunday :

<?php
$today = date('c');
$tomorrow = date('c', strtotime('+1 day'));

for($i=0;$i<=4;$i++){ 
    if(date('l',strtotime("+$i day")) != "Saturday" && date('l',strtotime("+$i day")) != "Sunday"){
        echo "<label>";
        if(date('c',strtotime("+$i day")) == $today){
            echo "Today";
        } elseif(date('c',strtotime("+$i days")) == $tomorrow){
            echo "Tomorrow";
        } else {
            echo date('l',strtotime("+$i day"));
            echo ", </label><label>";
            echo date('M d', strtotime("+$i day"));
        }
        echo "</label><br>";
    }
 }

?>

Output :