我想编写一个PHP函数,以特定的方式回显sql查询结果。
例子我有table 1
是10.000 rows
* 43 columns
:NO NAME Prof Date of birth1 John Teacher 19872 Nick Engineer 1976345
等等。基于No
是integer
(1-10.000),我想生成如下表:Name: JohnProf: TeacherDate of birth: 1987
Name: NickProf: EngineerDate of birth: 1976
到目前为止,我的代码:
`
$hostname = "";
$username = "";
$password = "";
$dbname = "db1";
//connection to the database
$con = mysqli_connect($hostname, $username, $password, $dbname)
or die("Unable to connect to MySQL");
mysqli_set_charset($con, 'utf8');
/*echo "Connected to db1 database <br>";
else
echo "Could not connect"; */
$query1 = "SELECT * FROM `table 1` WHERE CODE_NO = 1";
$query2 = "SHOW COLUMNS FROM table 1";
$result1 = mysqli_query($con, $query1);
$result2 = mysqli_query($con, $query2);
// Check result
// Useful for debugging.
if (!$result1) {
$message = 'Invalid query: ' . mysqli_error($con) . "\n";
$message .= 'Whole query: ' . $query1;
die($message);
}
echo "<table>"; // table tag in the HTML
while($row = mysqli_fetch_array($result1))
//Creates a loop to loop through results
{
echo
"<tr>
<th>CODE_NO:</th>
<td>" . $row['CODE_NO'] . "</td>
</tr>
<tr>
<th>FIRST_NAME:</th>
<td>" . $row['FIRST_NAME'] . "</td>
</tr>
<tr>
<th>SURNAME:</th>
<td>" . $row['SURNAME'] . "</td>
</tr>
<tr>
<th>Date of birth:</th>
<td>" . $row['DOB'] . "</td>
</tr>
<tr>
<th>Date of death:</th>
<td> " . $row['DOD'] . "</td>
</tr>";
}
echo "</table>"; //Close the table in HTML
?>`
是否可以对过程进行一次编码而无需对任何事物进行硬编码,因此可以根据需要将其重复多次?
编辑:
$x = 1;$query = "SELECT * FROM
个人WHERE CODE_NO = '$x'";$result = mysqli_query($con, $query);
最佳答案
如果我对您的理解正确,这将满足您的需求:
function generateTableFromResult($result) {
$html = "<table>";
while($row = mysqli_fetch_array($result)) {
foreach($row as $column => $value) {
$html.="<tr><th>".$column."</th><td>".$value."</td></tr>";
}
}
$html.="</table>";
return $html;
}
// usage:
// ...
$result1 = mysqli_query($con, $query1);
echo generateTableFromResult($result1);