我已經在html中創建了一個窗體,其中多個按鈕的命名方式與(sub1,2,3..so on)不同,但是將相同的值賦予所有窗體(value =「Update」)。我有一個問題,當任何按鈕被點擊時,它可以控制我所有的輸入框,所以我尋找解決這個問題的方法,並遇到了if..elseif語句的php腳本,但是當我試圖運行它時,只有前兩個if和elseif條件被編譯並且其餘條件被忽略。 下面是我的PHP代碼,所以請大家幫我弄清楚這個問題。我敢肯定,我的問題會幫助許多人太:)如果... elseif在超過2個條件下無法工作?
<?php
if($_POST['header']){
if(isset($_POST['header']) && trim($_POST['header']) != ""){
$fp = fopen("assets/config_header.php", 'w');
fwrite($fp, '<?php');
fwrite($fp, ' ');
fwrite($fp, '$header = "'.trim($_POST['header']).'";');
fwrite($fp, ' ');
fwrite($fp, '?>');
fclose($fp);
}
}
elseif($_POST['color']) {
if(isset($_POST['color']) && trim($_POST['color']) != ""){
$fp = fopen("assets/config_header5.php", 'w');
fwrite($fp, '<?php');
fwrite($fp, ' ');
fwrite($fp, '$color= "'.trim($_POST['color']).'";');
fwrite($fp, ' ');
fwrite($fp, '?>');
fclose($fp);
}
}
elseif($_POST['headline']) {
if(isset($_POST['headline']) && trim($_POST['headline']) != ""){
$fp = fopen("assets/config_header1.php", 'w');
fwrite($fp, '<?php');
fwrite($fp, ' ');
fwrite($fp, '$headline = "'.trim($_POST['headline']).'";');
fwrite($fp, ' ');
fwrite($fp, '?>');
fclose($fp);
}
}
elseif($_POST['color1']) {
if(isset($_POST['color1']) && trim($_POST['color1']) != ""){
$fp = fopen("assets/config_header6.php", 'w');
fwrite($fp, '<?php');
fwrite($fp, ' ');
fwrite($fp, '$color1= "'.trim($_POST['color1']).'";');
fwrite($fp, ' ');
fwrite($fp, '?>');
fclose($fp);
}
}
elseif($_POST['subhead']) {
if(isset($_POST['subhead']) && trim($_POST['subhead']) != ""){
$fp = fopen("assets/config_header2.php", 'w');
fwrite($fp, '<?php');
fwrite($fp, ' ');
fwrite($fp, '$subhead = "'.trim($_POST['subhead']).'";');
fwrite($fp, ' ');
fwrite($fp, '?>');
fclose($fp);
}
}
elseif($_POST['color2']) {
if(isset($_POST['color2']) && trim($_POST['color2']) != ""){
$fp = fopen("assets/config_header7.php", 'w');
fwrite($fp, '<?php');
fwrite($fp, ' ');
fwrite($fp, '$color2= "'.trim($_POST['color2']).'";');
fwrite($fp, ' ');
fwrite($fp, '?>');
fclose($fp);
}
}
elseif($_POST['prodname']) {
if(isset($_POST['prodname']) && trim($_POST['prodname']) != ""){
$fp = fopen("assets/config_header3.php", 'w');
fwrite($fp, '<?php');
fwrite($fp, ' ');
fwrite($fp, '$prodname= "'.trim($_POST['prodname']).'";');
fwrite($fp, ' ');
fwrite($fp, '?>');
fclose($fp);
}
}
elseif($_POST['vidid']) {
if(isset($_POST['vidid']) && trim($_POST['vidid']) != ""){
$fp = fopen("assets/config_header4.php", 'w');
fwrite($fp, '<?php');
fwrite($fp, ' ');
fwrite($fp, '$vidid= "'.trim($_POST['vidid']).'";');
fwrite($fp, ' ');
fwrite($fp, '?>');
fclose($fp);
}
}
?>
感謝提前:),是的,如果有執行該任務的任何較短的方式,那麼請不要讓我知道:)
你不需要'else if'當你關閉if時 –
你想檢查裏面的條件嗎? – Drudge
@Drudge不,我不希望它被嵌套,但要在其他如果梯子 – Codeweb24