[Solved] PHP if statement with OR operators [closed]

Use || statement. <?php if($CURRENT_USER[‘selected_plan_id’] == ‘4’ || $CURRENT_USER[‘selected_plan_id’] == ‘5’ || $CURRENT_USER[‘selected_plan_id’] == ‘6’): ?> Or Use in_array(). $arr = Array (4,5,6); <?php if(in_array($CURRENT_USER[‘selected_plan_id’],$arr)): ?> 1 solved PHP if statement with OR operators [closed]

[Solved] php if elseif statement select wrong variable every time [closed]

You have to write if($_POST[‘postnummer’] == “7900” || $_POST[‘postnummer’] == “7950” || $_POST[‘postnummer’] == “7960”) { $region = “Nordjylland”; } elseif ($_POST[‘postnummer’] == “8654” || $_POST[‘postnummer’] == “8660” || $_POST[‘postnummer’] == “8680” || $_POST[‘postnummer’] == “8700”) { $region = “Midtjylland”; } solved php if elseif statement select wrong variable every time [closed]

[Solved] Php Calculating two functions [closed]

You can’t access $dine in your second function because it’s nowhere defined. The $dine from your first function is only a local variable. I would suggest this solution which uses the fact that calculate_dinein_price() also returns the value of $dine: public function calculate_dinein_total(){ $total_dine = 0.00; $total_dine = $total_dine + $this->calculate_dinein_price(); return $total_dine; } 8 … Read more

[Solved] Ajax code the auto refresh a php file and updates it’s vaules [closed]

In your load.php at the end: echo json_encode($loadout); In index.html <script type=”text/javascript” src=”https://ajax.googleapis.com/ajax/libs/jquery/1.9.0/jquery.min.js”></script> <script type=”text/javascript”> //Calling function repeatAjax(); function repeatAjax(){ jQuery.ajax({ type: “POST”, url: ‘load.php’, dataType: ‘json’, success: function(resp) { jQuery(‘#out1’).html(resp[0]); jQuery(‘#out2’).html(resp[1]); jQuery(‘#out3’).html(resp[2]); }, complete: function() { setTimeout(repeatAjax,1000); //After completion of request, time to redo it after a second } }); } </script> 5 solved … Read more

[Solved] Writing an IF statement [closed]

Something like this should work: <div class=”<?php echo $this->getSize(); ?>”> <?php if (!empty($this->getBlockLink())) : ?> <a href=”https://stackoverflow.com/questions/25530264/<?php echo $this->getBlockLink(); ?>”> <?php endif; ?> <span class=”grid grid–full”> <span class=”grid__item <?php echo $this->getTextPosition() . ‘ ‘ . $this->getTextWidth(); ?>”> <?php echo $this->getBlockFormattedContent(); ?> </span> <img src=”<?php echo $this->getImage(); ?>”> </span> <?php if (!empty($this->getBlockLink())) : ?> </a> <?php … Read more

[Solved] How to echo a JS variable to php? [duplicate]

You can not pass the JavaScript variable up to PHP to index the $list array. What you can do is to pass the whole array down to JavaScript as JSON encoded and then index into that. See: Convert php array to Javascript 0 solved How to echo a JS variable to php? [duplicate]

[Solved] PHP, combine two arrays into a new array [closed]

Sounds to me like you’re looking for array_merge, or array_merge_recursive Perhaps a better fit would be: $result = array(); for($i=0, $j= count($arr1);$i<$j;$i++) {//standard loop over array $result[$i] = array_merge($arr1[$i], $arr2[$i]); } That should give you what you need. But please, do look into various array_* functions, there’s 79 of them in total, odds are that … Read more

[Solved] in array the image extension get variables

Supposing you have array of file names: $files = array(‘file1.jpg’, ‘file2.PNG’, ‘file3.txt’); You might want to filter images only like this: $extensions = array(‘jpg’, ‘jpeg’, ‘png’); $images = array_filter($files, function($file) use ($extensions) { $pos = strrpos($file, ‘.’); $extension = strtolower(substr($file, $pos + 1)); return in_array($extension, $extensions); }); (According to posts on php.net, pathinfo is a … Read more