if /else 内的 foreach

Question

目前我有这个

// display all address with their associated phone numbers
    foreach ($data->address as $address) {
      if ($address->fullAddress) echo $address->fullAddress;
      if ($address->phones->PhoneNumber) echo $address->phones->PhoneNumber);
    }

，但如果不满足上述子句中的语句，我需要添加这个作为替代方案

// display all phone numbers if there is not an address
    if ($data->relatedPhone) echo $data->relatedPhone;

如果没有满足任何子句，我根本不希望显示任何内容。那么我怎样才能将两者结合起来呢？

Answer 1

这是一个快速简单的解决方案：

$addrfound=false;
foreach ($data->address as $address) {
  if ($address->fullAddress) { echo $address->fullAddress; $addrfound=true; }
  if ($address->phones->PhoneNumber) { echo $address->phones->PhoneNumber); $addrfound=true; } 
}

if (!$addrfound)
    if ($data->relatedPhone) echo $data->relatedPhone;

Answer 2

您可以执行 else if 而不是另一个 if：

if (foo) {
  echo "bar";
} else if (foo2) {
  echo "bar2";
} else {
  echo "foobar";
}

仅当所有其他 if 语句时才运行最后一个 else评估为假。

所以对于你的情况，我会尝试这个：

// display all address with their associated phone numbers
    foreach ($data->address as $address) {
      if ($address->fullAddress) {
        echo $address->fullAddress;
      } else if ($address->phones->PhoneNumber) {
        echo $address->phones->PhoneNumber;
      } else if ($data->relatedPhone) {
        echo $data->relatedPhone;
      }
    }

Answer 3

您需要将逻辑构建为更容易处理的 if 语句：

// display all address with their associated phone numbers
foreach ($data->address as $address) 
{
    $hasAddress = (bool) $address->fullAddress;
    $hasPhone = (bool) $address->phones->PhoneNumber;
    $isAddress = ??; # add your decision here, I can't see you specified it in your question.

    if ($hasAddress) echo $address->fullAddress;
    if ($hasPhone) echo $address->phones->PhoneNumber;

    // display all phone numbers if there is not an address
    if (!$isAddress) {
        if ($data->relatedPhone) echo $data->relatedPhone;
    }
}

Answer 4

您可以创建一个标志来指示是否至少满足其中一个条件。

$flag = 0;
if (condition_1) {
  echo $output_1;
  $flag = 1;
}

if (condition_2) {
  echo $output_2;
  $flag = 1;
}

if (!$flag && condition_3) {
  echo $output_3;
}