我正在通过表单上传图片和文件。但上传后名称没有变化。在第一个函数调用时,如果名称是profilePic,那么即使我在$ config ['file_name']参数中更改名称,此文件名在每次调用时都保持相同。
函数调用:
package main
import (
"fmt"
)
func main() {
_inChannel := _inListener(generator())
for val := range _inChannel {
fmt.Print(val, " -- ")
}
}
func generator() chan int { // NEED TO CALCULATE FACTORIAL FOR 100 NUMBERS
ch := make(chan int) // CREATE CHANNEL TO INPUT NUMBERS
go func() {
for i := 1; i <= 100; i++ {
ch <- i
}
close(ch) // CLOSE CHANNEL WHEN ALL NUMBERS HAVE BEEN WRITTEM
}()
return ch
}
func _inListener(ch chan int) chan int {
rec := make(chan int) // CHANNEL RECEIVED FROM GENERATOR
go func() {
for num := range ch { // RECEIVE THE INPUT NUMBERS FROM GENERATOR
result := factorial(num) // RESULT IS A NEW CHANNEL CREATED
rec <- <-result // MERGE INTO A SINGLE CHANNEL; rec
close(result)
}
close(rec)
}()
return rec // RETURN THE DEDICATED CHANNEL TO RECEIVE ALL OUTPUTS
}
func factorial(n int) chan int {
ch := make(chan int) // MAKE A NEW CHANNEL TO OUTPUT THE RESULT
// OF FACTORIAL
total := 1
for i := n; i > 0; i-- {
total *= i
}
ch <- total
return ch // RETURN THE CHANNEL HAVING THE FACTORIAL CALCULATED
}
//上传功能
function getAndSaveEmployeeDetails() {
$resume = 'resume';
$resume = $this -> do_upload($resume, $employeeID);
$profilePic = 'profilePic';
$profilePic = $this -> do_upload($profilePic, $employeeID);
$cnic = 'cnicScannedImage';
$cnicScannedImage = $this -> do_upload($cnic, $employeeID); }
请查看以下链接中的图片,了解行为
These are the file names and it is caching the first name passed in function call
答案 0 :(得分:3)
使用
在upload
的第一行加载getAndSaveEmployeeDetails()
库
$this->load->library('upload');
没有传递任何配置。
然后在定义配置数组后的do_upload($uploadedField, $employeeID)
函数中,不要使用$this->load->library('upload', $config);
但请使用$this->upload->initialize($config);