Question

如何显示各自数据的列名。

$sql="SELECT * FROM <Tablename> WHERE domain_name='google.com'";
$result=mysqli_query($conn,$sql);
$row=mysqli_fetch_array($result,MYSQLI_NUM);
echo "<table>";
for ($i=0; $i <sizeof($row) ; $i++) { 
    echo "<tr>";
    echo "<td>";
    echo mysql_field_name($row,$i); // its not working.
    echo $row[$i]."<br/>";
    echo "</td>";
    echo "</tr>";
}
echo "</table>";

请帮忙。

Answer 1

您可以使用fetch_field_direct：

for ($i=0; $i <sizeof($row) ; $i++) { 
   ...
   $finfo = $result->fetch_field_direct($i);
   echo $finfo->name;
   ...
}

Answer 2

试试此代码

if ($result->num_rows > 0) {
    while($row = mysqli_fetch_array($result,MYSQLI_NUM)) {
    echo "<tr>";
    echo "<td>";
    echo $row[0]."<br/>";
    echo "</td>";
    echo "</tr>";
    }
    echo "</table>";
} else {
    echo "0 results";
}
$conn->close();

Answer 3

使用while(true) { size_t bytes_avail; ioctl(fd, FIONREAD, &bytes_avail); if(bytes_avail == 0) break; int n = read (fd, buf, (sizeof buf)-1); if(n<0) { //error handling fprintf(stderr, "Error while receiving message: %s\n", strerror(errno)); return; } buf[n] = '\0'; // you can add a check here to see if (length+n) < 1024 strcat(msg, buf); length += n; }获取列名称。

mysqli_fetch_assoc

您将获得$sql='SELECT * FROM mytesttable WHERE username="as"'; $result=mysqli_query($conn,$sql); $row=mysqli_fetch_assoc($result); header('Content-type: application/json'); echo json_encode($row);。如果您需要在output as json response中输出，将来这可能会有所帮助。谢谢，希望这对其他人有帮助。

在MySQL中显示包含数据的列名

3 个答案: