[英]Echo inner join to $row
您好我有這個查詢來填充數據庫中的表。
$sql = "SELECT * FROM clientes";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<table class='table table-bordered table-striped'>";
echo "<thead>";
echo "<tr>";
echo "<th>#</th>";
echo "<th>Grupo</th>";
echo "<th>Grupo</th>";
echo "<th>Ação</th>";
echo "</tr>";
echo "</thead>";
echo "<tbody>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['id_cliente'] . "</td>";
echo "<td>" . $row['id_grupo'] . "</td>";
echo "<td>" . $row['cliente'] . "</td>";
echo "<td>";
echo "<a href='read.php?id=". $row['id_cliente'] ."' title='Ver Registo' data-toggle='tooltip'><span class='mdi mdi-magnify'></span></a>";
echo "<a href='update.php?id=". $row['id_cliente'] ."' title='Actualizar Registo' data-toggle='tooltip'><span class='mdi mdi-pencil'></span></a>";
echo "<a href='delete.php?id=". $row['id_cliente'] ."' title='Apagar Registo' data-toggle='tooltip'><span class='mdi mdi-sync'></span></a>";
echo "</td>";
echo "</tr>";
}
echo "</tbody>";
echo "</table>";
// Free result set
mysqli_free_result($result);
} else{
echo "<p class='lead'><em>Sem registos</em></p>";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
// Close connection
mysqli_close($link);
?>
echo $ row ['id_grupo']中的值是數字的,但與另一個表相關。
我知道查詢要在php myadmin內部運行,以便從兩個表中內部聯接值:
SELECT grupos.grupo
FROM clientes
INNER JOIN grupos
ON clientes.id_grupo = grupos.id_grupo
我如何在$ row ['id_grupo']上回應呢?
您可以在此處使用Alias作為列名,例如:
SELECT clientes.id_grupo as c_group, grupos.id_grupo as g_group
// rest of your query.
那么您可以在$row
以下列:
$row['c_group'] // value from clientes table
$row['g_group'] // value from groups table
旁注 : MySQL ALIASES可用於為列或表創建臨時名稱。
使用方法:
SELECT clientes.*, grupos.grupo
FROM clientes
INNER JOIN grupos
ON clientes.id_grupo = grupos.id_grupo
而是SELECT * FROM clientes
並且您在while
擁有$row['grupo']
的值
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