dongpu3347 2013-05-28 17:32
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将SQL表导出为CSV,不显示列名

So here is the code I currently have for exporting a specific table to .csv format. It works perfectly so far, as in it downloads the file and formats the data how I would like it. However it doesn't display the Column names

ex: It currently shows

1 Jones Matt

2 Smith John

3 Doe Jane

I would like it to show:

ID Last_Name First_Name

1 Jones Matt

2 Smith John

3 Doe Jane

<?php
header("Content-type: application/csv");
header("Content-Disposition: attachment; filename=applications.csv");
header("Pragma: no-cache");
header("Expires: 0");

ini_set('display_errors',1);
$private=1;
error_reporting(E_ALL ^ E_NOTICE);

mysql_connect("localhost", "user", "pass") or die(mysql_error());
mysql_select_db("db") or die(mysql_error());

$query = "SELECT * FROM applications";
$select_c = mysql_query($query) or die(mysql_error());

while ($row = mysql_fetch_array($select_c, MYSQL_ASSOC))
{
    $result.="{$row['ID']},";
    $result.="{$row['LAST_NAME']},";
    $result.="{$row['FIRST_NAME']},";
    $result.="{$row['ORGANIZATION']},";
    $result.="{$row['TITLE']},";
    $result.="
";

}
    echo $result;
?>
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4条回答 默认 最新

  • dongqiulei6805 2013-05-28 17:37
    关注

    Here's a quick and dirty way of adding the column names:

    <?php
    header("Content-type: application/csv");
    header("Content-Disposition: attachment; filename=applications.csv");
    header("Pragma: no-cache");
    header("Expires: 0");
    
    ini_set('display_errors',1);
    $private=1;
    error_reporting(E_ALL ^ E_NOTICE);
    
    mysql_connect("localhost", "user", "pass") or die(mysql_error());
    mysql_select_db("db") or die(mysql_error());
    
    $query = "SELECT * FROM applications";
    $select_c = mysql_query($query) or die(mysql_error());
    
    // -------------------- add the line below -----------------------------------
    $result="ID,LAST_NAME,FIRST_NAME,ORGANIZATION,TITLE
    ";
    // -------------------- add the line above -----------------------------------
    
    while ($row = mysql_fetch_array($select_c, MYSQL_ASSOC))
    {
        $result.="{$row['ID']},";
        $result.="{$row['LAST_NAME']},";
        $result.="{$row['FIRST_NAME']},";
        $result.="{$row['ORGANIZATION']},";
        $result.="{$row['TITLE']},";
        $result.="
    ";
    
    }
        echo $result;
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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