PHP / SQL - How to update a database field

Question

so I have in my website a table of clients with several columns such as name, age, etc, etc... and I have added a new column which is "Delete" and I added checkboxes, and there's also a button, what I want to do is, if I select one or more checkboxes, and then click on the button to delete, I want them to be deleted off the website but then on the database I want to a field to be updated from 0 to 1 whereas 0 means the client exists on both website and database and 1 means that it no longer exists on the website but it still exists on database.

I'm new in the area of programming but this is the code I got so far:

$id = $_POST['id'];

if(isset($_POST['option']) == 'check' && ($_POST['button']) == 'submit') {

$query = "UPDATE commande

SET traiter = 1

WHERE id = '%$id%'";

var_dump($query);

}

Now this doesn't work, he is not doing the update correctly, I select the checkbox and then click on the button and it doesn't delete from the website nor updates the database specific field from the right client based on its' id, and I have no idea why!

Can anyone help me out here please?

You're using LIKE syntax in a simple comparison. Remove the percent signs after WHERE. Also: obligatory Bobby Tables reference. — Linus Kleen
– Linus Kleen, Commented Dec 8, 2015 at 9:18
You're right @Epodax and that's the problem, I don't get it why it's not working right — Le Future
– Le Future, Commented Dec 8, 2015 at 9:35

Jyoti Sharma · Accepted Answer · 2015-12-08 09:41:15Z

2

You can use this by two ways:

1-If more than one id(s) are coming then :

$query = "UPDATE commande SET traiter = 1 WHERE id IN ($id)";

2-If only one id will update at a time then:

$query = "UPDATE commande SET traiter = 1 WHERE id = '$id'";

answered Dec 8, 2015 at 9:41

Jyoti Sharma

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Abhishek Sharma · Accepted Answer · 2015-12-08 09:18:53Z

0

if you use = remove %

$query = "UPDATE commande SET traiter = 1 WHERE id = '$id'";

or use LIKE with %

$query = "UPDATE commande SET traiter= 1 WHERE id LIKE '%$id%'";

answered Dec 8, 2015 at 9:18

Abhishek Sharma

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11 Comments

Maha Dev Over a year ago

Using 'like' with $id doesn't make sense

Le Future Over a year ago

Oh true, lack of attention of mines, already corrected that small detail but still doesn't work, I'm using var_dump to debug it, and when I execute the action I want to do I get this message from var_dump: string(56) "UPDATE commande SET traiter = 1 WHERE id LIKE '%%'"

Abhishek Sharma Over a year ago

run this query on UPDATE commande SET traiter = 1 WHERE id = '$id' MySQL and check working fine or not ???? @LeFuture

Maha Dev Over a year ago

But usually we use id as an primary key. I don't think id will be like 'abc001'

Sunny S.M Over a year ago

check first is variable $id = $_POST['id']; contain value or NOT.. debug your script... solution would be there

|

AkshayP · Accepted Answer · 2015-12-08 09:43:36Z

0

Multiple select checkbox for delete

Considering that $id is an array, You can useIN clause to replace many OR conditions.

You can use in clause in where condition like

$query = "UPDATE commande SET traiter = 1 WHERE id IN ($id)";

Reference

Update

You should treat it as an array like this,

 <input name="id[]" type="checkbox" value="1">
 <input name="id[]" type="checkbox" value="2">
 <input name="id[]" type="checkbox" value="3">
 <input name="id[]" type="checkbox" value="4">

Then only, you can take it as an array. So, $_POST['id'] will provide array of selected checkbox.

edited Dec 8, 2015 at 9:43

answered Dec 8, 2015 at 9:29

AkshayP

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12 Comments

Le Future Over a year ago

$id is not an array, I just created a php variable named $id = $_POST['id']; to say that $id is equal to database field 'id'

AkshayP Over a year ago

There are many checkbox, right ? Keep that each checkbox name same and post the data. then you will get array

Le Future Over a year ago

All the checkboxes already have the same name, I've already covered that, the problem here is directly related to the query

AkshayP Over a year ago

Dude, your multiple condition in query can be done as I shown in my answer. Checkbox name should be like this id[] for each checkbox. So, $_POST['id'] will provide array of selected checkbox.

Le Future Over a year ago

so basically you're saying, in checkbox I put name="id[]" is that it?

|

Samir Karmacharya · Accepted Answer · 2015-12-13 04:55:14Z

0

after you get id/s of delete record 

    $query = "UPDATE commande SET traiter = 0 WHERE id IN ($id)";//For multiple multiple



    $$query = "UPDATE commande SET traiter = 0 WHERE id = '$id'";//for single record
 Set traiter to '0' means deleted record and '1' for active record

edited Dec 13, 2015 at 4:55

answered Dec 8, 2015 at 9:46

Samir Karmacharya

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