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PHP將MYSQL導出為CSV-獲取列標題,無法獲取其余數據?

[英]PHP EXPORT MYSQL TO CSV - got column headers, can't get rest of data?

我似乎無法弄清楚為什么我的表數據不會出現在csv文件中。 我有一頁提交表單以開始導出過程。 使用excel打開文件時我得到正確的列標題...

謝謝您的幫助!!

if(isset ($_POST['submit'])) {
        $host = "";
        $user = "";
        $password = "";
        $db = "";
        $con = mysql_connect("$host", "$user", "$password") or die("Couldn't connect to server.");  
        $db = mysql_select_db("$db", $con) or die ("Couldn't select database.");
        $table = "wp_users";

    function exportMysqlToCsv($table,$filename = 'content.csv') {

    $csv_terminated = "\n";

    $csv_separator = ",";

    $csv_enclosed = '"';

    $csv_escaped = "\\";

    $query = "select * from $table";

    $result = mysql_query($query);

    $rowcount = mysql_num_fields($result);

    $data = '';

    for ($i = 0; $i < $rowcount; $i++) {

    $l = $csv_enclosed . str_replace($csv_enclosed, $csv_escaped . $csv_enclosed,

    mysql_field_name($result, $i)) . $csv_enclosed;

    $data .= $l;

    $data .= $csv_separator; }

    $output = trim(substr($data, 0, -1));

    $output .= $csv_terminated;

    while ($row = mysql_fetch_array($result)) {

    $data = '';

    for ($j = 0; $j < $fields_cnt; $j++) {

    if ($row[$j] == '0' || $row[$j] != '') {

    if ($csv_enclosed == '') { $data .= $row[$j]; }

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

    else { $data .= ''; }

    if ($j < $rowcount - 1) { $data .= $csv_separator; } }

    $output .= $schema_insert;

    $output .= $csv_terminated; }

    header("Cache-Control: must-revalidate, post-check=0, pre-check=0");

    header("Content-Length: " . strlen($output));

    header("Content-type: text/x-csv");

    //header("Content-type: text/csv");

    //header("Content-type: application/csv");

    header("Content-Disposition: attachment; filename=$filename");

    echo $output;

    exit;

    } 

    exportMysqlToCsv($table);

    }

問題似乎是您正在將CSV字符串分配給$data變量,但沒有將其附加到$output 您需要這樣的一行:

$output .= $data;

之前:

$output .= $csv_terminated;

其次,我真的建議您修復代碼結構。 以目前的方式閱讀代碼非常困難。 閱讀PSR標准以獲取更多信息(或者有大量的體面標准)。

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