如何使用php仅显示全名
我设置了 MySQL 服务器，我正在尝试使用 Arduino 和以太网屏蔽从 SQL 服务器获取数据。问题是我无法从输出中删除无用的全文。
我不太了解 php，所以我不能尝试很多东西。
<html>
<head>
<title>Try Session JSON</title>
</head>
<body>
<?php
$dbusername = "root";
$dbpassword = "";
$server = "localhost";
$dbconnect = mysqli_connect($server, $dbusername, $dbpassword);
$dbselect = mysqli_select_db($dbconnect, "test");
$sql="SELECT full_name FROM test.eattandance WHERE id=1";
$records=mysqli_query($dbconnect,$sql);
$json_array=array();
while($row=mysqli_fetch_assoc($records))
{
$json_array[]=$row;
}
/*echo '<pre>';
print_r($json_array);
echo '</pre>';*/
echo json_encode($json_array);
?>
</body>
</html>
我希望 tryjson.php 的输出为 A1，但实际输出为[{"full_name":"A1"}].
HUX布斯1回答
-
慕桂英3389331
这是你想要的吗？ <html> <head> <title>Try Session JSON</title> </head> <body> <?php $dbusername = "root"; $dbpassword = ""; $server = "localhost"; $dbconnect = mysqli_connect($server, $dbusername, $dbpassword); $dbselect = mysqli_select_db($dbconnect, "test"); $sql="SELECT full_name FROM test.eattandance WHERE id=1"; $records=mysqli_query($dbconnect,$sql); $json_array=array(); while($row=mysqli_fetch_assoc($records)) { $json_array[]=$row; echo $row['full_name']; } ?> </body> </html>
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