私は本当に奇妙な状況にあります、私はこのように私のjavascript関数を呼び出しています...
window.top.window.stopUpload(<? echo $result; ?>,<? echo $file_name; ?>);
Javascript関数は次のようになります。
function stopUpload(success,filePath){
var result = '';
if (success == 1){
result = '<span class="msg">The file was uploaded successfully!<\/span><br/><br/>';
}
else {
result = '<span class="emsg">There was an error during file upload!<\/span><br/><br/>';
}
document.getElementById('f1_upload_process').style.visibility = 'hidden';
document.getElementById('f1_upload_form').innerHTML = result + '<input name="image_file" type="file" class="browse" /><input type="submit" name="submit_button" value="Upload" class="browse"/>';
document.getElementById('f1_upload_form').style.visibility = 'visible';
return true;
}
上記のコードはstopUpload
関数を実行しません。
しかし、私がこれが好きなら、
window.top.window.stopUpload(<? echo $result; ?>);
そしてこのようなJavaScript、
function stopUpload(success){
var result = '';
if (success == 1){
result = '<span class="msg">The file was uploaded successfully!<\/span><br/><br/>';
}
else {
result = '<span class="emsg">There was an error during file upload!<\/span><br/><br/>';
}
document.getElementById('f1_upload_process').style.visibility = 'hidden';
document.getElementById('f1_upload_form').innerHTML = result + '<input name="image_file" type="file" class="browse" /><input type="submit" name="submit_button" value="Upload" class="browse"/>';
document.getElementById('f1_upload_form').style.visibility = 'visible';
return true;
}
1つのパラメータで動作します!
質問
2つではなく1つのパラメータで機能するのはなぜですか?'hello'
の代わりに通常の文字列を送信しようとし$file_name
ましたが、それでも呼び出されません。