CODEWITHSUNDEEP
phpyii
nywuctuk2
nywuctuk2

Reputation: 93

Yii: model->save - unpredictable saving

I have some view for every objects of array ($this->receiveList)and there are 2 buttons(2 forms with input submit) for every objects too. If i trying to change status value of someone of objects(models) - only one(1st) object will be changed(saved new value for status) by all 4 buttons. BUT ONLY 2 BUTTONS READY FOR THAT AND 2 for another object.
WHY ITS HAPPENING Guys?
Example img: http://s017.radikal.ru/i441/1510/c7/974e00e3f8be.jpg

$this->receiveList = Invite::model()->findAll( $criteria2 );    // array of 2 objects

for ( $key=0; $key <= count($this->receiveList) - 1; $key++ ) {
    // yes
    if ( isset($_POST['formInviteYes']['sendRequest']) ) {
        unset($_POST['formInviteYes']['sendRequest']);

        $this->receiveList[$key]->status = 1;
        $this->receiveList[$key]->save(false);

        $this->refresh();
    }
    // no
    if ( isset($_POST['formInviteNo']['sendRequest']) ) {
        unset($_POST['formInviteNo']['sendRequest']);

        $this->receiveList[$key]->status = 0;
        $this->receiveList[$key]->save(false);


        $this->refresh();
    }

    print('<form action="" method="post" id="">');
        print('<input type="submit" name="formInviteYes[sendRequest]" value="Согласиться"" class="inviteAnswer">');
    print('</form>');

    print('<form action="" method="post" id="">');
        print('<input type="submit" name="formInviteNo[sendRequest]" value="Отказаться" class="inviteAnswer">');
    print('</form>');
}

Upvotes: 0

Views: 105

Answers (2)

aalgogiver
aalgogiver

Reputation: 196

You need to name submit buttons to contain info about object it's related to

        print('<input type="submit" name="formInviteNo[' . $invite->id . '][sendRequest]" value="Отказаться" class="inviteAnswer">');

And then you could implement your logic in this way:

$this->receiveList = Invite::model()->findAll( $criteria2 );    // array of 2 objects

foreach ($this->receiveList as $invite) {
    // yes
    if ( isset($_POST['formInviteYes'][$invite->id]['sendRequest']) ) {
        unset($_POST['formInviteYes'][$invite->id]['sendRequest']);

        $invite->status = 1;
        $invite->save(false);

        $this->refresh();
    }
    // no
    if ( isset($_POST['formInviteNo'][$invite->id]['sendRequest']) ) {
        unset($_POST['formInviteNo'][$invite->id]['sendRequest']);

        $invite->status = 0;
        $invite->save(false);

        $this->refresh();
    }

print('<form action="" method="post" id="">');
    print('<input type="submit" name="formInviteYes[' . $invite->id . '][sendRequest]" value="Согласиться"" class="inviteAnswer">');
print('</form>');

print('<form action="" method="post" id="">');
    print('<input type="submit" name="formInviteNo[' . $invite->id .'][sendRequest]" value="Отказаться" class="inviteAnswer">');
print('</form>');

}

Upvotes: 1

ScaisEdge
ScaisEdge

Reputation: 133400

You print the form whitout action related and same id

Try assign proper action to each form and also a proper id 

print('<form action="action1.php" method="post" id="id1">');
    print('<input type="submit" name="formInviteYes[sendRequest]" value="Согласиться"" class="inviteAnswer">');
print('</form>');

print('<form action="action2.php" method="post" id="2">');
    print('<input type="submit" name="formInviteNo[sendRequest]" value="Отказаться" class="inviteAnswer">');
print('</form>');

Upvotes: 1

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