Question

我是PHP和MySQL的新手。我知道其他语言，但是对于此项目，它是用PHP完成的。我在网上找到了一些需要调整的代码，因此它可以完成我的任务。我正在尝试从MySQL数据库导出数据，将其显示在屏幕上，并提供下载到excel的选项。该代码可以运行并且可以运行，但是它不会将来自SQL数据库的数据仅放在表列上。

<?php
$conn = new mysqli('localhost', 'root', '');   
mysqli_select_db($conn, 'hcap');   

$data = mysqli_query($conn,"SELECT `Org_ID`,'Org_Name', 'Org_Address', 'Org_Address2', 'Org_City', 'Org_State', 'Org_Zip', 'Org_County',
'Org_Website', 'Org_Phone', 'Org_fax', 'Org_Email'
FROM `organization`"); 

if(isset($_POST["ExportType"])) {
    switch($_POST["ExportType"]) {
        case "export-to-excel" :
            // Submission from
            $filename = $_POST["ExportType"] . ".xls";       
            header("Content-Type: application/vnd.ms-excel");
            header("Content-Disposition: attachment; filename=\"$filename\"");
            ExportFile($data);
            //$_POST["ExportType"] = '';
            exit();
        default :
            die("Unknown action : ".$_POST["action"]);
            break;
    }
}
function ExportFile($records) {
    $heading = false;
    if(!empty($records))
        foreach($records as $row) {
            if(!$heading) {
                // display field/column names as a first row
                echo implode("\t", array_keys($row)) . "\n";
                $heading = true;
            }
            echo implode("\t", array_values($row)) . "\n";
        }
        exit;
}
?>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/css/bootstrap.min.css">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/css/bootstrap-theme.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/js/bootstrap.min.js"></script>
<title>HCAP Organization Report</title>

<div><h3>HCAP Organization Report</h1></div>
<div> 
    <div id="container" >
        <div class="col-sm-6 pull-left">
            <div class="well well-sm col-sm-12">
                <b id='project-capacity-count-lable'><?php echo count($data);?></b> records found.
                <div class="btn-group pull-right">
                    <button type="button" class="btn btn-info">Action</button>
                    <button type="button" class="btn btn-info dropdown-toggle" data-toggle="dropdown">
                        <span class="caret"></span>
                        <span class="sr-only">Toggle Dropdown</span>
                    </button>
                    <ul class="dropdown-menu" role="menu" id="export-menu">
                        <li id="export-to-excel"><a href="#">Export to excel</a></li>
                        <li class="divider"></li>
                        <li><a href="#">Other</a></li>
                    </ul>
                </div>
            </div>
            <form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="post" id="export-form">
                <input type="hidden" value='' id='hidden-type' name='ExportType'/>
            </form>
            <table id="" class="table table-striped table-bordered">
                <tr>
                    <th>Organization ID</th>
                    <th>Organization Name</th>
                    <th>Organization Address</th>
                    <th>Organization Address 2</th>
                    <th>City </th>
                    <th>State </th>
                    <th>Zip Code </th>
                    <th>County </th>
                    <th>Website </th>
                    <th>Phone Number </th>
                    <th>Fax Number </th>
                    <th>E-Mail Address </th>
                </tr>
                <tbody>
                    <?php foreach($data as $row):?>
                        <tr>
                            <td><?php echo $row ['Org_ID']?></td>
                            <td><?php echo $row ['Org_Name']?></td>
                            <td><?php echo $row ['Org_Address']?></td>
                            <td><?php echo $row ['Org_Address2']?></td>
                            <td><?php echo $row ['Org_City']?></td>
                            <td><?php echo $row ['Org_State']?></td>
                            <td><?php echo $row ['Org_Zip']?></td>
                            <td><?php echo $row ['Org_County']?></td>
                            <td><?php echo $row ['Org_Website']?></td>
                            <td><?php echo $row ['Org_Phone']?></td>
                            <td><?php echo $row ['Org_fax']?></td>
                            <td><?php echo $row ['Org_Email']?></td>
                        </tr>
                    <?php endforeach; ?>
                </tbody>
            </table>
        </div>
    </div>
</div>
</body>   
<script  type="text/javascript">
  $(document).ready(function() {
    jQuery('#export-menu li').bind("click", function() {
      var target = $(this).attr('id');
      switch(target) {
        case 'export-to-excel' :
          $('#hidden-type').val(target);
          //alert($('#hidden-type').val());
          $('#export-form').submit();
          $('#hidden-type').val('');
          break
      }
    });
  });
</script>

Answer 1

您使用单引号而不是反引号。这就是为什么您在希望看到数据库值的地方看到文字列名称的原因。

改为使用此：

start <= x < finish

或者只是不使用任何引号（因为您的查询是不必要的-无需担心保留关键字或特殊字符）。

我在本地主机上测试了此建议的真实性和准确性。

使用PHP从MySQL导出数据

1 个答案: