Question

此代码仅将最后一行放入json;

<?php
$mysql_db_hostname = "localhost";
$mysql_db_user = "db_user";
$mysql_db_password = "db_password";
$mysql_db_database = "db_name";


$con = @mysqli_connect($mysql_db_hostname, $mysql_db_user, $mysql_db_password,
 $mysql_db_database);

if (!$con) {
 trigger_error('Could not connect to MySQL: ' . mysqli_connect_error());
}
$var = array();
 $sql = "SELECT id,name FROM db_table";
$result = mysqli_query($con, $sql);

while($obj = mysqli_fetch_object($result)) {
$var = $obj;
}
echo '{"users":'.json_encode($var).'}';
?>

我可以请求帮助来纠正我的代码中的错误吗？

Answer 1

将其替换为：

while($obj = mysqli_fetch_object($result)) {
    $var[] = $obj;
}
$data = ['users' => $var];
echo json_encode($data);

这将使$ var成为一个对象数组，否则你将覆盖$ var值。

Answer 2

在$var

之后使用数组符号

$var = array();
  while($obj = mysqli_fetch_object($result)) {
  $var[]  = $obj;
}

document.getElementById('YOURFORMID').reset();

我的代码只获得了JSON的最后一行

2 个答案: