Compare MySQL data with another table - php

Question

This code below echoes out 8 names from tableOne, but I want to compare those names with 8 names in another table. I want to compare the rows echoed in $row['weight'] with tableTwo, and if the results don't match, then add a <span class="strike"> </span> to the result echoed in $row['weight'].
How do I go about adding an if/else to $row['weight'] compare each name with the names in another table?
$result = mysqli_query($con,"SELECT * FROM tableOne LIMIT 0, 8");
$i = 1;
while($row = mysqli_fetch_array($result)) {
echo $i. " - " . $row['weight'] . '<br>';
$i++;
}

Answer 1

Here is some simple code to get you started:
$result = mysqli_query($con,"SELECT * FROM tableOne LIMIT 0, 8");
$result2 = mysqli_query($con,"SELECT * FROM tableTwo LIMIT 0, 8");
$i = 1;
while($row = mysqli_fetch_array($result)) {
$value1 = $row['weight'];
$row2 = mysqli_fetch_array($result2);
$value2 = $row2['weight'];
echo $i . " - table 1: ";
echo $value1;
echo ", table 2: - ";
if ($value2 != $value1) {
echo '<span class="strike">$value2</span>';
} else {
echo $value2;
}
echo '<br>';
$i++;
}
You can make the code smarter, to handle cases where there aren't 8 values to compare, and to display the values in an HTML table too, but hopefully this can get you started.

Answer 2

Try this:
$sql="select * from tableone to left join tabletwo tw on to.weight=tw.weight ";
This query will return all the rows in tableone which matches and empty rows for the ones where the weight dont match.
So in your php code:
while($row = mysqli_fetch_array($result)) {
if(empty($row['weight'])){
echo '<span class="strike">$row[weight]</span>';
}
}
This would work with any dynamic number of records in the tables.

Answer 3

How about keeping the values ($row[]) in one array($tableOne) and doing the same thing for table two ($tableTwo) and then perform your comparison in foreach()? (For the sake of simplicity, if you don't want any joins)