PHP if函数具有多个结果[重复]

Question

我正在尝试根据它所在的部门返回不同的属性状态。有销售和租赁部门，每个部门都有一个数字系统表示属性的状态。

我尝试使用 if 函数来查询每个属性的部门和状态，并为每个状态返回不同的文本字符串，但是所有属性都标记为“暂停”（即第一个选项）

代码如下。这不是特别优雅，抱歉。

function prop_status( $dept = null, $availability = null) {
if ( $dept = "Lettings" and $availability = "1") {
  return "On hold";
}
if ( $dept = "Lettings" and $availability = "2") {
  return "For rent";
}
if ( $dept = "Lettings" and $availability = "3") {
  return "References pending";
}
if ( $dept = "Lettings" and $availability = "4") {
  return "Let agreed";
}
if ( $dept = "Lettings" and $availability = "5") {
  return "Let";
}
if ( $dept = "Sales" and $availability = "1") {
  return "On hold";
}
if ( $dept = "Sales" and $availability = "2") {
  return "For sale";
}
if ( $dept = "Sales" and $availability = "3") {
  return "Under offer";
}
if ( $dept = "Sales" and $availability = "4") {
  return "Sold STC";
}
if ( $dept = "Sales" and $availability = "5") {
  return "Sold";
}
if ( $dept = "Sales" and $availability = "7") {
  return "Withdrawn";
}

非常感谢您对这项工作的任何帮助。

Answer 1

您应该将所有= 更改为==

function prop_status( $dept = null, $availability = null) {
if ( $dept == "Lettings" and $availability == "1") {
  return "On hold";
}
if ( $dept == "Lettings" and $availability == "2") {
  return "For rent";
}
if ( $dept == "Lettings" and $availability == "3") {
  return "References pending";
}
if ( $dept == "Lettings" and $availability == "4") {
  return "Let agreed";
}
if ( $dept == "Lettings" and $availability == "5") {
  return "Let";
}
if ( $dept == "Sales" and $availability == "1") {
  return "On hold";
}
if ( $dept == "Sales" and $availability == "2") {
  return "For sale";
}
if ( $dept == "Sales" and $availability == "3") {
  return "Under offer";
}
if ( $dept == "Sales" and $availability == "4") {
  return "Sold STC";
}
if ( $dept == "Sales" and $availability == "5") {
  return "Sold";
}
if ( $dept == "Sales" and $availability == "7") {
  return "Withdrawn";
}

查看php中的比较运算符here

$a == $b 如果 $a 等于 $b，则返回 TRUE

$a === $b 如果 $a 等于 $b，则返回 TRUE，并且它们属于同一类型。