我是一个laravel的乞丐。我想获取表格中输入的信息。表单字段名称为name
,email
和contact.
,我试图像核心php中的$_POST['name'] ,$_POST['email']
一样获取
{{form::open(array('route'=>'employees.store'))}}
controller store()函数:这将从表单中获取所有输入。我想要个人投入。
public function store()
{
$input=Input::all();
employee::create($input)
}
答案 0 :(得分:0)
store()
但是public function store(Request $request)
不应该像这样收到Request对象吗?
public function store(Request $request)
{
$input = $request->all();
$e = new Employee();
$e->name = $input['name'];
$e->email = $input['email'];
$e->contact = $input['contact'];
...
$e->save();
}
然后你可以这样做
void replaceSpace(char *s) {
int spaces = 0;
for (int i = 0; i < strlen(s); i++) {
if (s[i] == ' ') {
spaces++;
}
}
// new string that includes overwriting space, and two additional chars
int newLen = strlen(s) + spaces * 2;
s[newLen] = '\0';
for (int i = strlen(s) - 1; i >= 0; i--) {
if (s[i] == ' ') {
s[newLen - 1] = '0';
s[newLen - 2] = '2';
s[newLen - 3] = '%';
newLen -= 3;
}
else {
s[newLen - 1] = s[i];
--newLen;
}
}
}
char test[] = "rep lace Spac e";
replaceSpace(test);
cout << test << endl; //rep%20lace%20Spac%20e