我在PHP编程方面不是很好,并且还在学习它。 这是我的问题,我需要显示来自两个不同表的行的结果,研究过的东西并尝试但都失败了。
希望有人能用我的代码行给出一些建议。
$query = "SELECT tblparent.*, tblchild.* FROM tblparent, tblchild* FROM tblparent";
$num_results = $result->num_rows;
$result = $mysqli->query( $query );
if( $num_results ){
echo "<center><table border='1' id='members'>";
echo "<tr>";
echo "<th>Parent ID</th>";
echo "<th>Parent Firstname</th>";
echo "<th>Parent Lastname</th>";
echo "<th>Parent Middlename</th>";
echo "<th>Child ID</th>";
echo "<th>Child Firstname</th>";
echo "<th>Child Middlename</th>";
echo "<th>Child Lastname</th>";
echo "<th>Action</th>";
echo "</tr>";
while( $row = $result->fetch_assoc() ){
extract($row);
echo "<tr>";
echo "<td>{$Parent_ID}</td>";
echo "<td>{$PFname}</td>";
echo "<td>{$PLname}</td>";
echo "<td>{$PMname}</td>";
echo "<td>{$Child_ID}</td>";
echo "<td>{$CFname}</td>";
echo "<td>{$CMname}</td>";
echo "<td>{$CLname}</td>";
echo "<td>";
echo "<a href='#' onclick='delete_mem( {$Parent_ID} );'>Delete</a>";
echo "</td>";
echo "</tr>";
}
echo "</table>";
}
else{
echo "No records found.";
}
$result->free();
$mysqli->close();
我看到两个错误:
你需要调整以下代码,我假设tblparent有一个id,tblchild与tblparent id有关系,因为parent_id:
$result