I have looked for the question, to see if something appears about it, but I think I have not seen it, if so, I apologize in advance.
The point is the following, I have a query in mysql where in theory I should show 2 values, otherwise, do not show anything.
In my database there are types of properties (apartments or apartments, houses, etc.) and types of operations (sale, rent, etc.).
I have a search engine on my database, where what I do is receive the information from my form and this should give me what I need.
For example if in my form $ type = 'apartment' and $ operation = 'sale', I should show only that, but the fact is that it shows me all the properties for sale, even if they are not apartments. This is the sentence:
$busqueda = $conexion->query("SELECT * FROM inmuebles WHERE operacion = '$operacion' AND tipo = '$tipo' ORDER BY precio_dolar ASC LIMIT 100");
I know that in PHP there is the operator == that would force the search to be exactly the same as the string that is being requested, my question is punctual, there is something similar in Mysql, because I have not found it in the manuals.
!!! APOLOGIZE the whole Community !!! but I already found my fault. It has nothing to do with the sentence, but with the way you code my script.
What happens is the following, I am trying to create a search engine where there are 8 options, in theory, these options are filters and should work, but what happens.
If you select 1 option only, it shows all the properties with that single option and so the options are added until the 8 selected options are completed.
What happened, this:
if(!empty($operacion)) {
##el codigo
}
elseif (!empty($operacion) AND !empty($tipo)) {
##3el codigo
}
In both cases $ operation is not an empty string "", therefore both sentences are going to be executed, but if you deleted the first if, the statement executes perfectly.
Forgive my ignorance,: - (