Find text with like - mysql

Question

I want to find text in a MySql database
$result = mysqli_query($con,"SELECT * FROM blog
WHERE text_post LIKE = 't%';
")
or die(mysqli_error());
while($row = mysqli_fetch_array($result)) {
echo $row['titol_post'] . "<br>";
}
It gives me the error:
mysqli_error() expects exactly 1 parameter, 0 given...
If I substitute
WHERE text_post LIKE = 't%';
by
WHERE text_post = 'test';
It works well. I do not undestand. Why the LIKE does not work?

Answer 1

Here like after = is probelm please replace
with
WHERE text_post LIKE = 't%';
to
WHERE text_post LIKE 't%';

Answer 2

mysqli_error(), like all other mysqli_xx() functions, requires that you pass the connection variable to it, so that it knows which DB connection you want to know the last error for.
...
or die(mysqli_error($con));
...
Once you've done this, you will get a more useful error message that will help you diagnose the problem with your SQL code.
When you do get the real SQL error message, you will find that the problem is with the = sign.
The reason for this is that LIKE is an operator in the same way as =. You can only use one operator here, so if you're using LIKE, then you don't need = as well.
Your SQL code would therefore change to look like this:
SELECT * FROM blog
WHERE text_post LIKE 't%

Answer 3

I just changed like = to like and it works
SELECT * FROM blog WHERE text_post LIKE 't%'"

Answer 4

Try to pass conn in the error function:
or die(mysqli_error($con));
The MySQL syntax says:
string mysqli_error(mysqli link);
Returns the last error message for the most recent MySQLi function
call that can succeed or fail.
Also you need to remove the = sign after the LIKE keyword
$result = mysqli_query($con,"SELECT * FROM blog
WHERE text_post LIKE 't%';
")
or die(mysqli_error($con));

Answer 5

This is the right way to write sql query in mysql. Please replace your query with this.
$result = mysqli_query($con,"SELECT * FROM blog
WHERE text_post LIKE 't%';
")