[英]Returning results after Post PHP
I have a script that loops through 3 files. 我有一个遍历3个文件的脚本。 Each file posts data to a api and returns a result, either Accepted or Rejected in variable $result.
每个文件将数据发布到api并返回变量$ result中的Accepted或Rejected结果。
On a Accepted response the script stops running and echo out the result. 在“接受”响应上,脚本停止运行并回显结果。 On a Rejected it echo's out the rejected response and carries on to the next file.
在被拒绝时,它回显被拒绝的响应并继续到下一个文件。
The problem is : Lets say all 3 files give a Rejected reponse , it echo's out 3 rejected responses. 问题是:假设所有3个文件都给出了拒绝响应,它回显了3个拒绝响应。
How can i only echo out a single generic response if it does not get a Accepted Response. 如果没有收到“已接受的响应”,我如何仅回显单个通用响应。
foreach($getSeq as $key){
$fileName = $key->file;
include_once 'Lenders/' . $fileName;
if($result == 'Accepted'){
echo 'Accepted';
break;
}
if($result == 'Rejected'){
echo 'Rejected';
}
}
Try this, this is specifically for 3 reject response 试试这个,这是专门针对3次拒绝的响应
$failed = 0;
foreach($getSeq as $key){
$fileName = $key->file;
include_once 'Lenders/' . $fileName;
if($result == 'Accepted'){
echo 'Accepted';
break;
}
if($result == 'Rejected'){
$failed++;
}
}
if($failed == 3) {
echo "Rejected";
}
You didn't said if what will happens to other ratio (1A:2R, 2A:1R) , so I just created it for 3 rejects 您没有说其他比率(1A:2R,2A:1R)会怎样,所以我只是为3次拒绝创建了它
you can do it with array() like below: 您可以使用array()如下所示:
<?php
$resultArr = array(); // suppose we insert below Accepted = 1 and Rejected = 0
foreach($getSeq as $key){
$fileName = $key->file;
include_once 'Lenders/' . $fileName;
if($result == 'Accepted'){
echo "Accepted";
$resultArr[] = 1;
break;
}
if($result == 'Rejected'){
$resultArr[] = 0;
}
}
if(!in_array(1, $resultArr)){
echo 'Rejected';
}
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