Check if there is a value in db with if - php

Question

I have to check if a value is present in two tables, through a left join, if there is a relationship write yes, otherwise no
every time I associate an evidence (tev_Evidenze) to the structure (tev_Tipi_accreditamento) the query should tell me for that structure there is evidence, only now at structure n6. no evidence is present and anyway I answer yes
code:
<?php
$CONTROLLA = mysqli_query($riskmanagement,
"SELECT * FROM tev_Tipi_accreditamento LEFT JOIN tev_Evidenze
ON tev_Tipi_accreditamento. ID_tipo_acc = tev_Evidenze.id_tipo_accreditamento
WHERE tev_Tipi_accreditamento.id_struttura = tev_Evidenze.id_struttura GROUP BY tev_Evidenze.id_struttura");
$EVIDENZE=mysqli_num_rows($CONTROLLA);
if($EVIDENZE==0) {
echo "SI";
}else{
echo "no";
}
?>
The cycle is correct I am sure that in the db, there are no values ​​present in both tables but my if does not seem to work
NOTE:
<?php
$query_string = "SELECT * FROM tev_Tipi_accreditamento LEFT JOIN tev_Evidenze
ON tev_Tipi_accreditamento. ID_tipo_acc = tev_Evidenze.id_tipo_accreditamento
WHERE tev_Tipi_accreditamento.id_struttura = tev_Evidenze.id_struttura GROUP BY tev_Evidenze.id_struttura";
$query = mysqli_query($riskmanagement, $query_string);
?>
<?php
while($row = mysqli_fetch_assoc($query)){ ?>
<?php echo $row['id_struttura'] ;
if($query_string==0) {
echo "SI";
}else{
echo "no";
}?>
<?php } ?>

Answer 1

You're counting ALL the matches, not the number of matches for each structure. You can use the following:
SELECT s.id_struttura, IF(COUNT(e.id_struttura) > 0, 'Yes', 'No') AS matches
FROM tev_Tipi_accreditamento AS s
LEFT JOIN tev_Evidenze AS e ON s.id_struttura = e.id_struttura AND s.ID_tipo_acc = e.id_tipo_accreditamento
GROUP BY s.id_struttura
This will return a table like:
1 Yes
2 No
3 No
4 Yes
for each structure ID.
DEMO
You can show the result with:
$CONTROLLA = mysqli_query($riskmanagement, "
SELECT s.id_struttura, IF(COUNT(e.id_struttura) > 0, 'Yes', 'No') AS matches
FROM tev_Tipi_accreditamento AS s
LEFT JOIN tev_Evidenze AS e ON s.id_struttura = e.id_struttura AND s.ID_tipo_acc = e.id_tipo_accreditamento
GROUP BY s.id_struttura") or die(mysqli_error($riskmanagement));
echo "<table>";
while ($row = mysqli_fetch_assoc($CONTROLLA)) {
echo "<tr><td>{$row['id_struttura']}</td><td>{$row['matches']}</td>
}
echo "</table>";