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Katılma tarihi: 13 May 2022
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Toxic (Drum Kit) WAV [April-2022]



 


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5 GB Sounds + 1.8 GB Spatial Music + 350+ Patches, 150+ Bonus Files, 230+ Soundbanks All In One Kit + This is A MUST HAVE KitQ: JSON returns a warning message I try to use the following php code to return JSON format to get the results from my mysql table, but it doesn't work. I get a warning message. Is it possible that the json_encode cannot get the results, how can I return the results properly? Thank you. A: You can just do this to get your result set as JSON in the first place. $result = mysql_query("SELECT * FROM Book"); $data = array(); while($row = mysql_fetch_assoc($result)) { $data[] = $row; } echo json_encode($data); I'm not sure why you're getting a warning message though. Q: Switch a variable while looping through results of multiple queries I want to make a button that when pressed queries a db and then displays the results. In the code below I display all the results and then switch through them by passing the index of the results through the variable $e, $i, $j etc. I thought of an if statement that checks if the variable $e is equal to 0 then pass a variable to the next iteration, however I want to do it in a switch statement as I need to do this for multiple iterations. Any thoughts?

 

 

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Toxic (Drum Kit) WAV [April-2022]
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