【发布时间】：2021-04-06 06:05:21
【问题描述】：
我正在处理yii2。我正在尝试在我的控制器中获取 submit button 名称。以下是我的看法
<div class="form-group">
<a class="btn btn-default" onclick="window.history.back()" href="javascript:"><i
class="fa fa-close"></i>
Cancel</a>
<?php if ($model->isNewRecord && in_array(Yii::$app->user->identity->access_level,['admin','manager'])) { ?>
<?= Html::submitButton('Add' ,['class' => 'btn btn-success','name'=>'a']) ?>
<?= Html::submitButton('Add & Send Quotation' ,['class' => 'btn btn-info','name'=>'s']) ?>
<?php } elseif (!$model->isNewRecord && in_array(Yii::$app->user->identity->access_level,['admin'])) {?>
<?= Html::submitButton('Update', ['class' =>'btn btn-warning']) ?>
<?php }elseif (!$model->isNewRecord && in_array(Yii::$app->user->identity->access_level,['editor'])){?>
<?php
if ($model->reqStatus==\common\models\Accountant::$request_status_titles[0]
||$model->reqStatus==\common\models\Accountant::$request_status_titles[2]
)
{
echo Html::submitButton('Verified', ['class' =>'btn btn-success','name'=>'v']);
echo Html::submitButton('Not-Verified', ['class' =>'btn btn-danger','name'=>'n']);
}
?>
<?php } elseif (!$model->isNewRecord && in_array(Yii::$app->user->identity->access_level,['manager'])){?>
<?php
if ($model->reqStatus==\common\models\Accountant::$request_status_titles[3])
{
echo Html::submitButton('Signed',['class' =>'btn btn-primary','name'=>'g']);
}
elseif ($model->reqStatus==\common\models\Accountant::$request_status_titles[4])
{
echo Html::submitButton('Follow-Up',['class' =>'btn btn-primary','name'=>'r']);
}
else{
echo Html::submitButton('Send Quotation',['class' =>'btn btn-primary','name'=>'q']);
}
?>
<?php } ?>
</div>
在上面的代码中有多个submit button，每个name。在我的控制器中，我想获得 Html::submitButton('Verified', ['class' =>'btn btn-success','name'=>'v']); 这个和 echo Html::submitButton('Not-Verified', ['class' =>'btn btn-danger','name'=>'n']); 这个按钮的名称。
if (isset($_POST['v'])) //verified
{
$model->load(Yii::$app->request->post());
$model->reqStatus = Accountant::$request_status_titles[1];
$model->callVerified = 1;
$model->clientRef = 'CL-'.uniqid();
if($model->save())
{
return $this->redirect(['view', 'id' => $model->id]);
}
return $this->render('update', [
'model' => $model,
]);
}
elseif (isset($_POST['n']))//not-verified
{
$model->load(Yii::$app->request->post());
$model->reqStatus = Accountant::$request_status_titles[2];
if($model->save())
{
return $this->redirect(['view', 'id' => $model->id]);
}
return $this->render('update', [
'model' => $model,
]);
}
else
{
if ($model->load(Yii::$app->request->post()) && $model->save()) {
return $this->redirect(['view', 'id' => $model->id]);
}
return $this->render('update', [
'model' => $model,
]);
}
当我执行它时。我可以看到verified 和not verified 按钮。但是当我点击其中任何一个时，它不会转到 if (isset($_POST['v'])) 或 esleif (isset($_POST['n']))part 而是点击 else 部分。我正在另一个控制器上做同样的事情，它工作得很好。
更新 1
根据scaisEdge 的评论，我尝试过var_dump($_POST)，下面是结果
array(2) {
["_csrf-backend"]=> string(88) "xS7Ctb2C677HKQJjoBL2SPznK3IHVGUDSn7MGhMs9SigHov36PSI7J9OewzkSJ0AzpBYN0A_KTIlEIZfd1qiYQ=="
["Bookkeeping"]=> array(43) {
["name"]=> string(5) "Laeeq"
["conNumb"]=> string(11) "03464394600"
["email"]=> string(15) "laeeq@gmail.com"
["location"]=> string(12) "Amber Valley"
["reqConcerns"]=> string(22) "UK_limited_partnership"
["workStart"]=> string(21) "PROJECT_START_2_MONTH"
["businessName"]=> string(27) "Retail_consumer_merchandise"
["reqPurpose"]=> string(40) "Select between several bookkeeper quotes"
["bookAssign"]=> string(16) "One_time_project"
["bookkRequire"]=> string(32) "Only local bookkeepers can quote"
["bookWork"]=> string(15) "Annual accounts"
["companyName"]=> string(22) "NONPROFIT_ORGANIZATION"
["turnOver"]=> string(17) "£300,000-600,000"
["info"]=> string(7) "Testing"
["remarks"]=> string(15) "Client Verified"
["annualConf"]=> string(0) ""
["personalTaxRetHrmc"]=> string(0) ""
["mPaySlip"]=> string(0) ""
["mRtiPenRet"]=> string(0) ""
["annualAcc"]=> string(0) ""
["annualCorpTaxRet"]=> string(0) ""
["annualCorpTaxRetHrmc"]=> string(0) ""
["qVatRet"]=> string(0) ""
["qBookMtd"]=> string(0) ""
["insuranceHmrc"]=> string(0) ""
["corpTaxPlan"]=> string(0) ""
["persTaxPlan"]=> string(0) ""
["businessAdvice"]=> string(0) ""
["fedTax"]=> string(0) ""
["provTax"]=> string(0) ""
["invPerM"]=> string(0) ""
["invPerMPrice"]=> string(0) ""
["add1"]=> string(0) ""
["add2"]=> string(0) ""
["add3"]=> string(0) ""
["add4"]=> string(0) ""
["add5"]=> string(0) ""
["add6"]=> string(0) ""
["add7"]=> string(0) ""
["add8"]=> string(0) ""
["add9"]=> string(0) ""
["add10"]=> string(0) ""
["remarks2"]=> string(0) ""
}
}
上述结果中没有v字符串。
我一定错过了一些我不知道的东西。
任何帮助将不胜感激。
【问题讨论】：
-
尝试在控制器中使用 var_dump($_POST) 检查 $_POST .. 的真实内容
-
@scaisEdge 你可以查看我的
update 1 -
@scaisEdge 字符串
v不存在 -
但是你有名字
n.. ? .如果您只有两者之一，这对您的代码是正确的..一个项目只能有一个名称.. -
@scaisEdge 如果我点击
verified按钮，那么传递的名称应该是v，如果我点击not verified按钮，那么传递的名称应该是n。在这两种情况下，它都没有发生。
标签： php controller yii2 yii2-advanced-app