doutan1905 2017-06-23 19:45
浏览 63

尝试使用jQuery从数据库中获取信息

I am VERY new to jQuery, and I have been trying to get information from my SQL database whenever the user chooses a new option in a datalist form. This is the code:

<form>
<input list="chemicals" name="chemicalsearch">
  <datalist id="chemicals">

 <?php while ($info3=mysql_fetch_array($info2)) {
      echo "<option value='$info3[name]'>$info3[formel]</option>";
        }

?>
</datalist>
</form>

<script type="text/javascript">

$('#chemicals').on('change', function() {
    var record_id = $(this).val(); 
    var data = {
        'id': record_id
    };

    $.ajax({
        type: "GET",
        url: '/Chemistry%20Calculator/getchemical.php', 
        data: data,
        success: function(response) {
            document.getElementById('formel').innerHTML = data.formel;
        },
        error: function(jqXHR, textStatus, errorThrown) {
        }
    });
});
</script>

And then the PHP file:

<?php

include_once 'connect.php';

$info="SELECT * from chemicals where name='$id'";
$info2=mysql_query($info) or die("Wrong link. This page does not exist.");
$info3=mysql_fetch_array($info2);

$name = $info3['name'];
$formel = $info3['formel'];
$massa = $info3['molmassa'];

$array = array($name, $formel, $massa);

$data = json_encode($array);

echo $data;

?>

Please be patient with me here, as I have never used jQuery before. And yes, I am aware of the fact that I'm using the old MySQL syntax, I will change that as soon as I get this working. Any help would be greatly appreciated :)

  • 写回答

2条回答 默认 最新

  • doudu5498 2017-06-23 19:49
    关注

    You are not writing what exactly is the problem, either from the side of php or javaScript. But anyway, correct your syntax to php regarding the variables. Like below:

    <?php while ($info3=mysql_fetch_array($info2)) {
      echo '<option value="'.$info3[name].'">'.$info3[formel].'</option>';
    }?>
    

    and later on your query...$info='SELECT * from chemicals where name="'.$id.'"'

    评论

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