Laravel request->input('') 或 old('') 在刀片中总是返回 1

Question

I want to get the url data into my input value.

Example:

• the url is http://example.com/user?username=example
• and my html input code is as below
• <input value="{{ request()->input('username') or old('username') }}">
• but the result always return "1" and not "example"

It seems like the "or" blade helper acts as conditional operator.

How can I get the username value from url into that input and still get the default value as old('username')?

Answer 1

Try with

<input value="{{ request()->input('username', old('username')) }}">

It will retrieve username as an input if it exists or default to the old value. You can also give a default value to the old helper.

Answer 2

这种方式更简单：

<input value="{{ (old('username')?old('username'):'') }}">

Answer 3

很简单的方法。

<input value="{{ request()->input('username') ?? old('username') }}">

Answer 4

As per official documentation 8.x

We use the helper request

The request function returns the current request instance or obtains an input field's value from the current request:

$request = request();

$value = request('key', $default);

the value of request is an array you can simply retrieve your input using the input key as follow

$username = request()->username; //for http://example.com/user?username=example