How to insert data in SQLite table using php?

Question

I am working on a php code as shown below which scans the directory ($src_dir) and list all mp4 files.

$src_dir = ('\\\ABCD-ST-001\Audio_Test\podcast\incoming_folder'); 

$mp4_files = preg_grep('~\.(mp4)$~', scandir($src_dir));

print_r(array_values($mp4_files));  // LineA

Here is the O/P obtained from Line#A:

Array ( [0] => 36031P.mp4 [1] => hello.mp4 )

I am getting two values at Index 0 and Index 1. The 1st is 36031P.mp4 and 2nd is hello.mp4

After that I have wrote a script which insert data in table Podcast_Export.

$db->exec("INSERT INTO Podcast_Export ('House#', 'Status') VALUES ('array_values($mp4_files)', 'Go')");   /* have to place this question on SO*/

On running the above script, I am getting the following data inside the table which is not I want.

array_values(Array) Go
array_values(Array) Go

Problem Statement:

I am wondering what changes I should make in the script above so that it inserts the following data inside the table:

36031   Go
hello   Go

@KIKOSoftware I have used the following code to make a connection with SQLite databse class MyDB extends SQLite3 { function __construct() { $this->open('database/Podcast.db'); } } $db = new MyDB(); if(!$db) { echo $db->lastErrorMsg(); } else { echo "Opened database successfully<br>"; } — flash
– flash, Commented Jun 19, 2019 at 15:10
I don't know who downvoted my question. I showed my effort as well. — flash
– flash, Commented Jun 19, 2019 at 15:20
Most downvotes are a hit-and-run thing, so I doubt your comment will ever be read by the perpetrator. Just ignore downvotes, they're a fact of live, you can't please everybody. I'll upvote to compensate. Did you know that upvotes are free? — KIKO Software
– KIKO Software, Commented Jun 19, 2019 at 15:23

devpro · Accepted Answer · 2019-06-19 14:59:09Z

1

You are inserting a string value 'array_values($mp4_files)' , you just need to remove quotes around this value. something like:

$db->exec("INSERT INTO Podcast_Export ('House#', 'Status') VALUES (array_values($mp4_files), 'Go')");

But, if you delete the quotes around array_values then it will insert following array as a string Array ( [0] => 36031P.mp4 [1] => hello.mp4 )

so, its better to use loop here with $mp4_files array and insert your data.

Side Note:

I suggest you to use different name for House# column something like house_no.

edited Jun 19, 2019 at 14:59

answered Jun 19, 2019 at 14:52

devpro

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7 Comments

flash Over a year ago

I have used the following code in order to insert data in the database and it works.

foreach ($mp4_files as $value) { 			$db->exec("INSERT INTO Podcast_Export ('House#', 'Status') VALUES ('".$value."', 'Go')"); 		}

flash Over a year ago

The issue which is now is that when I refresh the page it calls the script again and it will insert in the table many times.

devpro Over a year ago

@flash: as we dont know from where you get this variable $src_dir i think u need to add one more query before insert , check either data available of not... this is another question

devpro Over a year ago

@flash: i suggest you to create one another quesiotn with more detail from where you getting data, and close this task , marking acceptence, ...

flash Over a year ago

I have edited my question. It will give you more information from where I am getting $src_dir

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