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如何使用 json php 和 mysql 显示数据库中的数据。我一直在尝试，但我未能获胜，但如果我不包含 jquery，它会以一种有趣的方式显示。我将不胜感激任何帮助。贝娄是我到目前为止所尝试的。

<?php 
require_once('dbbox.php');
?>
<?php
$sql = "SELECT * FROM 
m1debtors INNER JOIN 
m1dtrans 
ON m1debtors.name = m1dtrans.user";
$result = mysql_query($sql, $con) or die(mysql_error($con));
while ($row = mysql_fetch_array($result)){
$accnum =  $row['accnum'];
$name = $row['name'];
$addr1 = $row['addr1'];
}
// The JSON standard MIME header.
header('Content-type: application/json');
$array = array('Name'=>$name, 'user'=>$user);
 echo  json_encode($array);
?>
Here is the html page to display data 
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script>
$(document).ready(function(){
$.getJSON("view.php",{$name});
});
</script>
4

1 回答 1

0

视图.php

<?php 
  require_once('dbbox.php');

  $sql = "SELECT * FROM m1debtors 
          INNER JOIN m1dtrans 
            ON m1debtors.name = m1dtrans.user";
  $result = mysql_query($sql, $con) or die(mysql_error($con));
  while ($row = mysql_fetch_array($result)){
    $accnum =  $row['accnum'];
    $name = $row['name'];
    $addr1 = $row['addr1'];
  }

  // The JSON standard MIME header.
  header('Content-type: application/json');
  $array = array('name'=>$name, 'user'=>$user);
   echo  json_encode($array);
?>

索引.php

<html>
  <head>
   <script src="http://code.jquery.com/jquery-latest.js"></script>
   <script> 
      $(document).ready(function(){
        $.getJSON('view.php', function(data) {
          $('#myJson').html(data.name + ' ' + data.user);
        });
      });

    </script>
  </head>
  <body>
    <div id="myJson"></div>
  </body>
</html>
于 2013-01-22T12:09:03.030 回答