Question

尝试生成要导出为excel文件的报表表单数据库，但是当数据库中有数据时，我一直得到0结果。任何人都可以看看代码，让我知道我错了什么。代码需要进入从4 +

运行php 5.4+的服务器

<?php
//set date
$date = date("m/d");

$course_report = $_POST['course_report'];

$course_info = explode('~',$course_report);
$course_info[0]; // course_id
$course_info[1]; // course_date


$select = "SELECT DISTINCT * FROM registrants WHERE (paid='Y' AND course_id = '$course_info[0]' AND course_date = '$course_info[1]')";
$result = mysqli_query($dbc, $select);

// first get the header row
while($row = mysqli_fetch_assoc($result)) {
    $course = $row['course'];
    $coursedate = $row['course_date'];

    $export = mysql_query($select);
    $fields = mysql_num_fields($export); 
}

// Get header

$fields = isset($_POST['fields']);
for ($i = 0; $i < $fields; $i++) {
    $header .= mysqli_field_name($export, $i) . "\t";
}


// Then get all the rows of data
$export = isset($_POST['export']);
while($row = mysqli_fetch_row($result)) {
    $line = '';
    foreach($row as $value) {                                            
        if ((!isset($value)) OR ($value == "")) {
            $value = "\t";
        } else {
            $value = str_replace('"', '""', $value);
            $value = '"' . $value . '"' . "\t";
        }
        $line .= $value;
    }
    $data .= trim($line)."\n";
}
$data = isset($_POST['data']);
$data = str_replace("\r","",$data); 

// Check for blank data set
if ($data == "") {
    $data = "\n(0) Records Found!\n";                        
} 
// end building the data

$header = isset($_POST['header']);
$course = isset($_POST['course']);
$file_name = $course . "_downloaded_" . $date;
header("Content-type: application/x-msdownload");
//header("Content-Disposition: attachment; filename=Registrar_101.txt");
header("Content-Disposition: attachment; filename=$file_name.xls");
header("Pragma: no-cache");
header("Expires: 0");
print "$header\n$data";
// } 

?>

Answer 1

这个说它只能得到一行，但它实际上会得到所有行。

// first get the header row
while($row = mysqli_fetch_assoc($result)) {
    $course = $row['course'];
    $coursedate = $row['course_date'];

    $export = mysql_query($select);
    $fields = mysql_num_fields($export); 
}

然后，对于每一行，您都会用mysql_query()莫名其妙地重新执行查询在这一点上，我已经停止了寻找，因为如果脚本的其余部分完全没有意义的话，你真的需要清理这里的逻辑。

Answer 2

我不确定你认为在“首先得到标题行”下发生了什么。查询正在从“注册人”表中检索记录。你认为“标题行”来自哪里？您是否认为您将使用列标题从查询中获取一行，即字段的名称？如果是这样，那不是它的工作原理。

然后在此循环中，您将为每条记录重新执行查询。为什么？那取得了什么成果？然后，您将获得此内部查询中字段数的计数，并将其放入$ fields中。但你永远不会使用这个值。在循环之后，你立即覆盖它。

你明白isset（）会返回true还是false？你似乎认为它返回一个字符串。

$ _ POST获取从网页上提交的表单上的值。你似乎在想它会从你的查询或其他东西给你价值。

我不确定你要完成的是什么，但我认为你的程序的核心应该是这样的：

<?php
//set date
$date = date("m/d");

$course_id = $_POST['course'];
$course_date = $_POST['course_date'];

// You really should use mysqli_real_escape_string to escape input strings,
// but let's skip that point for now

// Also, it's generally bad form to use "select *", as then you have no idea what
// fields you will get back. Better to list exactly what you want.

$select = "SELECT DISTINCT * FROM registrants WHERE (paid='Y' AND course_id = '$course_id' AND course_date = '$course_date')";
// This assume the db is already open as $dbc
$result = mysqli_query($dbc, $select);

// build header row
$header = '';
$field_count = mysqli_num_field($result);
for ($i = 0; $i < $field_count; $i++) {
  $finfo = mysqli_fetch_field_direct($result,$i);
  $name = $finfo->name;
  $header .= $name . "\t";
  // I'm not familiar with a mysqli_field_name function and couldn't find it in the docs.
  // Whatever.
}

$data = $header . "\n";

// Then get all the rows of data
while($row = mysqli_fetch_row($result)) {
  $line = '';
  foreach($row as $value) {                                            
    if (!isset($value) OR $value == "") {
      $value = "\t";
    } else {
      $value = str_replace('"', '""', $value);
      $value = '"' . $value . '"' . "\t";
  $line .= $value;
  }
  $data .= trim($line)."\n";
}
... and then whatever you need to do with this thing ...

试图从SQL数据库打印报告？

2 个答案: