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zofe / rapyd-laravel

deprecated rewritten in rapyd-livewire
MIT License
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Call to undefined method Zofe\Rapyd\DataFilter\DataFilter::with() #234

Closed sonispark closed 9 years ago

sonispark commented 9 years ago

Hi, I want to filter details from a table where its status is zero.I tried like this $filter = \DataFilter::source(new \App\category); $filter->add('id','ID', 'text'); $filter->add('description','Description','text'); $filter->add('created_at','Created date','daterange')->format('m/d/Y', 'en'); $filter->submit('search'); $filter->reset('reset'); $filter->build();

    $grid = \DataGrid::source($filter::with('parent'));
    $grid->add('id','ID', true)->style("width:100px");
    $grid->add('title','Title');
    $grid->add('description','Description');

......

Here i am getting the error Call to undefined method Zofe\Rapyd\DataFilter\DataFilter::with().Please help me to solve this

tacone commented 9 years ago

try:

$filter = \DataFilter::source(\App\category::with('parent');
$grid = \DataGrid::source($filter);
sonispark commented 9 years ago

Hi,

I have a doubt.Can we use angular js with this... My problem is that

I have to dropdown box category and subcategory.The subcategory dropdown is filled if we select only the category from category dropdown box..How to do this in laravel...Hope your reply

Thank you Soni Joy

On Mon, Oct 12, 2015 at 5:22 PM, tacone notifications@github.com wrote:

try:

$filter = \DataFilter::source(\App\category::with('parent');$grid = \DataGrid::source($filter);

— Reply to this email directly or view it on GitHub https://github.com/zofe/rapyd-laravel/issues/234#issuecomment-147374319.

Warm regards, Soni Joy Software Engineer

Email: soni@sparksupport.com abey@sparksupport.com | * *Skype: spark.soni

http://www.sparksupport.com/ http://www.sparkmycloud.com/ https://www.facebook.com/sparksupport http://www.linkedin.com/company/244846 https://twitter.com/sparksupport3rd Floor, Leela Infopark, Phase -2,Kakanad, Kochi-30, Kerala, India Phone:+91 484 6561696 , Mobile:91-8129270421. Confidentiality Notice: Information in this e-mail is proprietary to SparkSupport. and is intended for use only by the addressed, and may contain information that is privileged, confidential or exempt from disclosure. If you are not the intended recipient, you are notified that any use of this information in any manner is strictly prohibited. Please delete this mail & notify us immediately at info@sparksupport.com

tacone commented 9 years ago

I'm sorry, this is not supported by Rapyd, at the moment.

sonispark commented 9 years ago

okey.. Thank you

On Tue, Oct 13, 2015 at 5:49 PM, tacone notifications@github.com wrote:

I'm sorry, this is not supported by Rapyd, at the moment.

— Reply to this email directly or view it on GitHub https://github.com/zofe/rapyd-laravel/issues/234#issuecomment-147698777.

Warm regards, Soni Joy Software Engineer

Email: soni@sparksupport.com abey@sparksupport.com | * *Skype: spark.soni

http://www.sparksupport.com/ http://www.sparkmycloud.com/ https://www.facebook.com/sparksupport http://www.linkedin.com/company/244846 https://twitter.com/sparksupport3rd Floor, Leela Infopark, Phase -2,Kakanad, Kochi-30, Kerala, India Phone:+91 484 6561696 , Mobile:91-8129270421. Confidentiality Notice: Information in this e-mail is proprietary to SparkSupport. and is intended for use only by the addressed, and may contain information that is privileged, confidential or exempt from disclosure. If you are not the intended recipient, you are notified that any use of this information in any manner is strictly prohibited. Please delete this mail & notify us immediately at info@sparksupport.com

sonispark commented 8 years ago

Hi

I am in a trouble. I have two table 'users' and 'user_details'. I am listing data from user_detail table with the user_id.Instead of listing user_id from user_details ,I want to display the user name from users table in the place of user_id from user_details table..

(I need Joining of two tables users and user_details as data source)

How to do this??

Hope your reply. Thanks in Advance

On Mon, Oct 12, 2015 at 5:22 PM, tacone notifications@github.com wrote:

try:

$filter = \DataFilter::source(\App\category::with('parent');$grid = \DataGrid::source($filter);

— Reply to this email directly or view it on GitHub https://github.com/zofe/rapyd-laravel/issues/234#issuecomment-147374319.

Warm regards, Soni Joy Software Engineer

Email: soni@sparksupport.com abey@sparksupport.com | * *Skype: spark.soni

http://www.sparksupport.com/ http://www.sparkmycloud.com/ https://www.facebook.com/sparksupport http://www.linkedin.com/company/244846 https://twitter.com/sparksupport3rd Floor, Leela Infopark, Phase -2,Kakanad, Kochi-30, Kerala, India Phone:+91 484 6561696 , Mobile:91-8129270421. Confidentiality Notice: Information in this e-mail is proprietary to SparkSupport. and is intended for use only by the addressed, and may contain information that is privileged, confidential or exempt from disclosure. If you are not the intended recipient, you are notified that any use of this information in any manner is strictly prohibited. Please delete this mail & notify us immediately at info@sparksupport.com