發送到服務器時添加一些字符的Base64字符串

Question

我正在開發一個應用程序，並且正在將base64編碼的圖像字符串發送到服務器。 字符串是正確的，當我從XCode調試器窗口復制字符串並從PHP生成圖像時，它將生成正確的有效圖像。 但是，當我通過Web服務發送相同的字符串時，它會在字符串中添加一些額外的字符，並且圖像會損壞。

如何從字符串中刪除這些多余的字符。

這是來自XCode調試器的圖像的base64字符串：

/9j/4AAQSkZJRgABAQAASABIAAD/4QBYRXhpZgAATU0AKgAAAAgAAgESAAMAAAAB
AAEAAIdpAAQAAAABAAAAJgAAAAAAA6ABAAMAAAABAAEAAKACAAQAAAABAAABd6AD
AAQAAAABAAABHgAAAAD/7QA4UGhvdG9zaG9wIDMuMAA4QklNBAQAAAAAAAA4QklN
BCUAAAAAABDUHYzZjwCyBOmACZjs+EJ+/8AAEQgBHgF3AwEiAAIRAQMRAf/EAB8A
AAEFAQEBAQEBAAAAAAAAAAABAgMEBQYHCAkKC//EALUQAAIBAwMCBAMFBQQEAAAB
fQECAwAEEQUSITFBBhNRYQcicRQygZGhCCNCscEVUtHwJDNicoIJChYXGBkaJSYn
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這是從服務器回顯的字符串：

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Answer 1

如注釋中所述，每64個字符插入一個“ rn”。 另外，“ +”將替換為“”。 讓我們扭轉一下。
用輸出字符串定義$s ：

$chars = str_split($s);
for($i = 64; $i < strlen($s); $i+=66) {
    unset($chars[$i]);
    unset($chars[$i+1]);
}
$new_s = str_replace(" ", "+", implode($chars));

for($i = 0; $i < strlen($new_s); $i++){
    if($i > 0 && $i % 64 == 0)
        echo "<br />";
    echo $new_s[$i];
}

演示： http : //sandbox.onlinephpfunctions.com/code/33ef917c4403560c469d8e9eea2ae88b4c6b732e
差異： https ： //www.diffchecker.com/ajAv578C