Question

我试图拉出一个php变量（$ purpose）并在选项select上显示它。我所拥有的代码（见下文）只有在我拿出元素属性ng-model =＆＃34;目的＆＃34;时才有效。如果我将ng-model留在了位置，则选项选择显示目的，这表示php echoinng没有通过。它似乎回应了php变量并使用angularjs相互冲突。有人有这个问题吗？有解决方法吗？

<select id="purpose" class="form-control custom-select" name="purpose" value="<?php if(isset($purpose) or !empty($purpose)) {echo $purpose;}?>" ng-model="purpose" autocomplete="on" required/>
  <option <?php if($purpose == "") echo 'selected';?> value="">Purpose</option>
  <option <?php if($purpose == "1030") echo 'selected';?> value="1030">1Exchange</option>
  <option <?php if($purpose == "1040") echo 'selected';?> value="1040">Non-exchange</option>
  <option <?php if($purpose == "1050") echo 'selected';?> value="1050">Financing</option>
</select>

Answer 1

使用 ng-model时不要使用值属性。只需初始化变量目的＆amp;为它分配PHP变量。如下所示：

<select ng-init="purpose = <?php if(isset($purpose) or !empty($purpose)) {echo $purpose;}?>" id="purpose" class="form-control custom-select" name="purpose" ng-model="purpose" autocomplete="on" required/>
      <option <?php if($purpose == "") echo 'selected';?> ng-value="">Purpose</option>
      <option <?php if($purpose == "1030") echo 'selected';?> ng-value="1030">1Exchange</option>
      <option <?php if($purpose == "1040") echo 'selected';?> ng-value="1040">Non-exchange</option>
      <option <?php if($purpose == "1050") echo 'selected';?> ng-value="1050">Financing</option>
    </select>

Answer 2

虽然您可以使用此样式使其工作，但建议使用JSON将数据从后端填充到角度。

<?php
$purpose_arr = array(
  "1030" => "1Exchange",
  "1040" => "Non-exchange",
  "1050" => "Financing"
);

echo "<script> var purpose='" + json_encode($purpose_arr) + "'; </script>"
?>

html:
<select id="purpose" class="form-control custom-select" name="purpose" ng-model="purpose" autocomplete="on" required
    ng-options="value as key for (key, value) in purpose" />
    <option value="">Purpose</option>
</select>

controller:
$scope.purpose = purpose;

angularjs干扰回显php变量

2 个答案: