codeignitor中的日期問題php

Question

我正在嘗試將$datestring分配給DAY的WDEK - 例如， 週日，週一等，因爲%w值是數字 - 這意味着1日，2是週二等等

這是一個codeignitor項目，我的代碼運行的第一個條件語句是否是真還是假！請幫忙。

 public function data() 
{ 

$datestring = "%w"; 
$time = time(); 

echo mdate($datestring, $time); 


if ($datastring = "1") { 

    echo "monday"; 
} 
else if ($datastring = "2") { 

    echo "tuesday"; 
} 
else if ($datastring = "3") { 

    echo "wednesday"; 
}  

else if ($datastring = "4") { 

    echo "thursday"; 
} 

else if ($datastring = "5") { 

    echo "friday"; 
} 
else if ($datastring = "6") { 

    echo "saturday"; 
} 

else if ($datastring = "7") { 

    echo "sunday"; 
} 
return; 
    } 

Answer 1

要比較你需要雙==，因爲=是分配的。並將「1」賦值給變量$ datastring始終爲真，因此總是首先如果爲真。將其更改爲：

if ($datastring == "1") { 

    echo "monday"; 
} 
else if ($datastring == "2") { 

    echo "tuesday"; 
} 
else if ($datastring == "3") { 

    echo "wednesday"; 
}  

else if ($datastring == "4") { 

    echo "thursday"; 
} 

else if ($datastring == "5") { 

    echo "friday"; 
} 
else if ($datastring == "6") { 

    echo "saturday"; 
} 
else if ($datastring == "7") { 

    echo "sunday"; 
} 

編輯：我還添加了一個else if語句星期天。您可以使用這個簡單的else也因爲沒有其他選擇留（只要$ datastring只能含量的數字1到七人。

爲了使它更清潔，我會用switch像

switch ($datastring) { 

    case 1: 
    $day = "monday"; 
    break; 
    case 2: 
    $day = "tuesday"; 
    break; 
    case 3: 
    $day = "wednesday"; 
    break; 
    case 4: 
    $day = "thursday"; 
    break; 
    case 5: 
    $day = "friday"; 
    break; 
    case 6: 
    $day = "saturday"; 
    break; 
    case 7: 
    $day = "sunday"; 
    break; 
    default: 
    $day = "unknown day"; 
    break;  
} 

echo $day;