Question

我在尝试使下面的语句按照我想要的方式工作时遇到了麻烦。

如果未填写至少一个文本字段，我正在尝试显示订单表单的错误消息。下面是我的PHP代码片段。 'cookieGiftBox'是用户可以选择的复选框的名称，如果选中它，它们必须在提供的文本字段中以所需的口味输入一定量的cookie。如果选中复选框但是没有填写文本字段，如何显示要显示的错误消息？

<?php
    if (isset($_POST['cookieGiftBox'])
        && (!isset($_POST['sugar']) 
            || ($_POST['chocolateChip'])
            || ($_POST['chocolateChipPecan']) 
            || ($_POST['whiteChocolateRaspberry'])
            || ($_POST['peanutChocolateChip']) 
            || ($_POST['peanutButter'])
            || ($_POST['tripleChocolateChip']) 
            || ($_POST['whiteChocolateChip'])
            || ($_POST['oatmealRaisin']) 
            || ($_POST['cinnamonSpice'])
            || ($_POST['candyChocolateChip']) 
            || ($_POST['butterscotchToffee'])
            || ($_POST['snickerdoodle']))) {
        $error.="<br />Please enter an Amount for Cookie Flavors";
    }
?>

Answer 1

＆安培;＆安培;优先于||，因此您必须使用括号来获得预期结果：

if checkbox is checked and not (sugar is checked or chocolatechip is checked)

实际上，要检查是否选择无，您可以执行以下操作： if checkbox is checked and sugar is not entered and chocolatechip is not entered...或同等的： if (isset($_POST['cookieGiftBox']) && !( isset($_POST['sugar']) || isset($_POST['chocolatechip']) ) ) { error ...}。如果您想了解更多信息，请搜索有关Boolean algebra。

的信息

更新：在你的例子和正确的语法中，我作为一个例子的句子是这样的（对于第一句，注意不是（！）和字段周围的括号，而不是复选框）：

if (isset($_POST['cookieGiftBox']) &&
    !isset($_POST['sugar']) &&
    !isset($_POST['chocolatechip'])
) { error...}

或者第二句，可能更容易理解（注意＆amp;＆amp;而不是||）：

(true && !false && !false)

如果选中giftbox复选框时，如果没有设置sugar，chocolatechips等字段，则使用ands确保它只是true（因此显示错误）。因此，如果选中该复选框，并且未设置任何字段，则它如下所示： (true && true && true)相当于true (true && !true && !false)，因此显示错误。如果选中该复选框并输入糖，则它将如下所示： (true && false && true)，等。到false，等。到[{"latitude":"40.651602","longitude":"-111.918734","title":"64439","content":"test"}][{"latitude":"41.330500","longitude":"-111.851600","title":"80283","content":"test"}][{"latitude":"40.624500","longitude":"-111.976500","title":"95964","content":"test"}][{"latitude":"40.715020","longitude":"-111.494717","title":"96995","content":"test"}][{"latitude":"40.664200","longitude":"-111.843295","title":"101724","content":"test"}][{"latitude":"40.513917","longitude":"-111.403612","title":"245697","content":"test"}][{"latitude":"40.509000","longitude":"-111.419919","title":"245716","content":"test"}][{"latitude":"40.473700","longitude":"-111.414499","title":"245742","content":"test"}][{"latitude":"40.509943","longitude":"-111.402999","title":"245792","content":"test"}][{"latitude":"40.512200","longitude":"-111.477263","title":"245874","content":"test"}][{"latitude":"40.526623","longitude":"-111.475199","title":"245904","content":"test"}][{"latitude":"40.530400","longitude":"-111.487500","title":"245905","content":"test"}][{"latitude":"40.526619","longitude":"-111.473531","title":"245921","content":"test"}][{"latitude":"40.517013","longitude":"-111.487827","title":"245926","content":"test"}][{"latitude":"40.519500","longitude":"-111.497780","title":"245940","content":"test"}][{"latitude":"40.511981","longitude":"-111.409799","title":"245944","content":"test"}][{"latitude":"39.373348","longitude":"-111.574996","title":"245946","content":"test"}][{"latitude":"40.514900","longitude":"-111.402590","title":"245982","content":"test"}][{"latitude":"40.513700","longitude":"-111.467712","title":"246093","content":"test"}][{"latitude":"40.515600","longitude":"-111.398192","title":"246124","content":"test"}]，因此没有显示错误。如果同时输入糖和巧克力碎片，也不会显示错误。

Answer 2

我有类似问题：

我尝试使用：

if (isset($this->context->controller->php_self) && ($this->context->controller->php_self != 'category') || ($this->context->controller->php_self != 'product') || ($this->context->controller->php_self != 'index')){ ... }

这不起作用，我改为：

if (isset($this->context->controller->php_self) && ($this->context->controller->php_self != 'category') && ($this->context->controller->php_self != 'product') && ($this->context->controller->php_self != 'index')){ ... }

我不知道这是对的吗？

编辑。这是不正确的，但现在我不知道如何解决它。

如果声明与＆amp;＆amp;和PHP中的多个OR条件

2 个答案: