2

我遇到了 SQL 查询问题...我有一个导出 CSV 文件的文件，该文件在我向查询添加约束之前一直有效。

这是我所拥有的

<?php
 $today = date("mdY"); 
 function exportMysqlToCsv($table,$filename = 'TodaysSubmissions.csv')
 {
$csv_terminated = "\n";
$csv_separator = ",";
$csv_enclosed = '"';
$csv_escaped = "\\";
$sql_query = "SELECT * FROM $table WHERE submitDate = '".$today."'";

我只想从今天开始导出所有记录，并且在我的表单中我对 $today 使用相同的日期格式，所以它应该匹配。

另外，当我更改为此时，它会完全按照我想要的方式创建 CSV 文件

$sql_query = "SELECT * FROM $table WHERE submitDate = 08142013";

那么当我在查询中调用 $today 时我做错了什么？我觉得我已经尝试了所有可能的方法。

编辑这里是整个文件

            $today = date("mdY"); 
        function exportMysqlToCsv($table,$filename = 'TodaysSubmissions.csv')
        {
            $csv_terminated = "\n";
            $csv_separator = ",";
            $csv_enclosed = '"';
            $csv_escaped = "\\";
            $sql_query = 'SELECT * FROM $table WHERE submitDate = "'.$today.'"';

            // Gets the data from the database
            $result = mysql_query($sql_query);
            $fields_cnt = mysql_num_fields($result);


            $schema_insert = '';

            for ($i = 0; $i < $fields_cnt; $i++)
            {
                $l = $csv_enclosed . str_replace($csv_enclosed, $csv_escaped . $csv_enclosed,
                    stripslashes(mysql_field_name($result, $i))) . $csv_enclosed;
                $schema_insert .= $l;
                $schema_insert .= $csv_separator;
            } // end for

            $out = trim(substr($schema_insert, 0, -1));
            $out .= $csv_terminated;

            // Format the data
            while ($row = mysql_fetch_array($result))
            {
                $schema_insert = '';
                for ($j = 0; $j < $fields_cnt; $j++)
                {
                    if ($row[$j] == '0' || $row[$j] != '')
                    {

                        if ($csv_enclosed == '')
                        {
                            $schema_insert .= $row[$j];
                        } else
                        {
                            $schema_insert .= $csv_enclosed .
                            str_replace($csv_enclosed, $csv_escaped . $csv_enclosed, $row[$j]) . $csv_enclosed;
                        }
                    } else
                    {
                        $schema_insert .= '';
                    }

                    if ($j < $fields_cnt - 1)
                    {
                        $schema_insert .= $csv_separator;
                    }
                } // end for

                $out .= $schema_insert;
                $out .= $csv_terminated;
            } // end while

            header("Cache-Control: must-revalidate, post-check=0, pre-check=0");
            header("Content-Length: " . strlen($out));
            // Output to browser with appropriate mime type, you choose ;)
            header("Content-type: text/x-csv");
            //header("Content-type: text/csv");
            //header("Content-type: application/csv");
            header("Content-Disposition: attachment; filename=$filename");
            echo $out;
            exit;
4

2 回答 2

4

如果您想让$today所有函数都可以使用该变量，并在函数外部声明它，则必须在函数内部声明它global：

<?php
$today = date("mdY"); 

...

function exportMysqlToCsv($table,$filename = 'TodaysSubmissions.csv')
{
  global $today;
  ...
  $sql_query = "SELECT * FROM $table WHERE submitDate = '".$today."'";

推荐阅读

您在“完整”代码中也有这个：

$sql_query = 'SELECT * FROM $table WHERE submitDate = "'.$today.'"';

双引号不起作用，这是肯定的。

于 2013-08-14T14:56:48.273 回答
0

$today未在您的函数中定义。将其移动到函数中。

 function exportMysqlToCsv($table,$filename = 'TodaysSubmissions.csv')
 {
    $today = date("mdY"); 
    $csv_terminated = "\n";
    $csv_separator = ",";
    $csv_enclosed = '"';
    $csv_escaped = "\\";
    $sql_query = "SELECT * FROM $table WHERE submitDate = '{$today}'";

使用双引号，您可以只添加变量而无需使用合并。

http://php.net/manual/en/language.variables.scope.php

于 2013-08-14T14:52:54.870 回答