How To Check If Input Type Button Is Pressed In Php?
Solution 1:
isset($_POST['Button'])
you just missed the capital B
. You need to match it the same as the input's name
attribute.
Solution 2:
you need to do that using ajax. so when a user clicks this button the page won't refresh.. there should be a page behind the seen does that for the user. whenever he clicks the button it accesses that page and upload whatever data you want...
edit: or you can just add a hidden input in the form when the button is clicked the hidden input value changes to i.e. true... then from php codes you can use the isset or other functions to validate whether the user clicked the button or not...
Solution 3:
In the example you have, the simplest approach would be to add an extra variable to the parameters passed in &button_press=true or something like that and then you would know that the button had been pressed when you are receiving the information
Solution 4:
'locrizak' answered right . Your button name is 'Button' and you tried to check presence of click on 'button' both are different . In such case if you are unsure of what is wrong , you may print the entire POST array using
print_r($_POST)
This will display all submitted values from form including button
Using isset() is the correct method to check whether a form element is present or not
Use
if(isset($_POST['Button'])){
//code block for insertion,validation etc //
}
Solution 5:
isset() function does not works with input type=button. so either we have to use input type=submit instead of button or some hidden type if we still want to use button.
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