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Is this SQL valid?

By : dminted
Date : October 17 2020, 11:12 PM
I hope this helps . No it isn't.
You need to specify the join columns for both tables, and you need to make sure you use a correct WHERE clause (which is missing from your query).
code :

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jQuery Validation conditional rule makes input never valid - valid always returns 0

By : Charles
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , I'm having a problem with jQuery validation conditional rules. , Turns out the correct syntax to add conditional rules are:
code :
TimeFrameAmount: { 
    required: { 
        depends : function () { 
            //check conditionality 

Error jquery validation plugin depends. Form valid when fields are not valid

By : Neeyo Jeey
Date : March 29 2020, 07:55 AM
I hope this helps . As I did not find any questions similar to mine around. The solution I found is the following:
Instead of using the depends method, use a function to establish if the field should or should not be required, including it in the conditions on which this operation should rely (i.e. if a checkbox is checked or not)
code :
$( "form[name=form]" ).validate({
    ignore: [],
    rules: {
        'form[owner]': {
            required: true,
        'form[bank_account]': {
            required: function(element){
                    return $('#my_select').is(':checked') && $('form_bank_account').val();
    messages : {
        'form[owner]' : {
             required: "The owner is necessary",
        'form[bank_account]' : {
            required: "This field must be filled out when switch on",

call_user_func_array() expects parameter 1 to be a valid callback, second array member is not valid method in

By : s.suriya narasimman
Date : March 29 2020, 07:55 AM
To fix the issue you can do What that error means is that $the_['function'] is an array, and so PHP is attempting to interpret $the_['function'][0] as an object and $the_['function'][1] as a method of that object, and a method of that name does not exist on the object. From the manual on callbacks:

com.google.firebase.FirebaseException: An internal error has occurred. [ API key not valid. Please pass a valid API key.

By : user3351841
Date : March 29 2020, 07:55 AM
Any of those help Hy i was also facing this issue.
Solution: change version of google services at project level gradle and sync your project and reinstall.

You must provide a valid Privacy Policy URL in order take your app Live. Go to App Details and make sure it is valid

By : akhiltata
Date : March 29 2020, 07:55 AM
Hope this helps Because you added Platform before you adding privacy policy Url and Terms of service url.
So please remove all the platform you have added under Settings-->Basic and then add your Privacy Policy Url Click on save changes. Now Enable your Status (Set To Live) again click on save changes. Now you can add your Platforms again and Click on Save Changes.
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