Count MySQL rows with values

Question

I want to count how many rows in my MySQL database have two certain values. My table is set up like this:

|---------------------|
|         ids         |
|---------------------|
|source_id | target_id|
|----------|----------|
|        2 |         6|
|        2 |         6|
|        3 |         4|
|---------------------|

I want to count how many rows have the source_id = 2 and target_id = 6 I tried this statement:

<?php
$prep_stmt = "SELECT source_id FROM ids WHERE source_id = 2 AND target_id = 6";
if (!$result = $mysqli->query($prep_stmt)) {
    die("Failed");
} else {
    $num_rows = $result->num_rows;
    echo $num_rows;
}
?>

However, the PHP file ceases to function after the third line.

Answer 1

SELECT COUNT(*) FROM ids WHERE source_id=2 AND target_id=6

Answer 2

Your code looks a bit weird. If you want to use prepared statements, that's working totally differentely:

<?php

$stmt = $mysqli->prepare("SELECT COUNT(*) FROM `ids` WHERE `source_id` = ? AND `target_id` = ?");
$stmt->bind_param("ii", $source_id, $target_id);
$stmt->execute();
$stmt->bind_result($count);
$stmt->fetch();
$stmt->close();
echo $count;

And without prepared statements.

<?php

echo $mysqli->query("SELECT COUNT(*) FROM `ids` WHERE `source_id` = 2 AND `target_id` = 6");

And as a last note, if you asign anything within a condition be sure to enclose it in brackets:

<?php

function fn() {
  return "something";
}

if (($foo = fn())) {
  // The condition is true if $foo isset, or in other words not null after the function was called.
}

if (!($foo = fn())) {}

if (($foo = fn()) === null) {}

// ...

Answer 3

SELECT COUNT(*) FROM ids WHERE source_id = 2 AND target_id = 6";

will give you the number of entries corresponding to what you want.

(it will give one row with 1 column, containing the number of lines corresponding to the where close)