Question

我对PHP不太陌生，试图从mysql数据库中获取数据并希望将其导出为excel。我也遵循this answer。

它有点旧，使用MYSQL连接代替，我希望它作为mysqli，我更改了如下代码

if (isset($_POST["submit"]))
{

$select = "SELECT * FROM `records`";

$export = mysqli_query($mysqli, $select); 
//$fields = mysql_num_rows($export); // thanks to Eric
$fields = mysqli_num_fields($export); // by KAOSFORGE
$col_title="";
$data="";
for ($i = 0; $i < $fields; $i++) {
    $col_title .= '<Cell ss:StyleID="2"><Data ss:Type="String">'.mysqli_fetch_field_direct($export, $i).'</Data></Cell>';
}

$col_title = '<Row>'.$col_title.'</Row>';

while($row = mysqli_fetch_row($export)) {
    $line = '';
    foreach($row as $value) {
        if ((!isset($value)) OR ($value == "")) {
            $value = '<Cell ss:StyleID="1"><Data ss:Type="String"></Data></Cell>\t';
        } else {
            $value = str_replace('"', '', $value);
            $value = '<Cell ss:StyleID="1"><Data ss:Type="String">' . $value . '</Data></Cell>\t';
        }
        $line .= $value;
    }
    $data .= trim("<Row>".$line."</Row>")."\n";
}

$data = str_replace("\r","",$data);

header("Content-Type: application/vnd.ms-excel;");
header("Content-Disposition: attachment; filename=export.xls");
header("Pragma: no-cache");
header("Expires: 0");

$xls_header = '<?xml version="1.0" encoding="utf-8"?>
<Workbook xmlns="urn:schemas-microsoft-com:office:spreadsheet" xmlns:x="urn:schemas-microsoft-com:office:excel" xmlns:ss="urn:schemas-microsoft-com:office:spreadsheet" xmlns:html="http://www.w3.org/TR/REC-html40">
<DocumentProperties xmlns="urn:schemas-microsoft-com:office:office">
<Author></Author>
<LastAuthor></LastAuthor>
<Company></Company>
</DocumentProperties>
<Styles>
<Style ss:ID="1">
<Alignment ss:Horizontal="Left"/>
</Style>
<Style ss:ID="2">
<Alignment ss:Horizontal="Left"/>
<Font ss:Bold="1"/>
</Style>

</Styles>
<Worksheet ss:Name="Export">
<Table>';

$xls_footer = '</Table>
<WorksheetOptions xmlns="urn:schemas-microsoft-com:office:excel">
<Selected/>
<FreezePanes/>
<FrozenNoSplit/>
<SplitHorizontal>1</SplitHorizontal>
<TopRowBottomPane>1</TopRowBottomPane>
</WorksheetOptions>
</Worksheet>
</Workbook>';

print $xls_header.$col_title.$data.$xls_footer;
exit;
}

但在第16行中却给了我错误，如下所示

$col_title .= '<Cell ss:StyleID="2"><Data ss:Type="String">'.mysqli_fetch_field_direct($export, $i).'</Data></Cell>';

让我知道是否有人可以帮助我解决问题。谢谢

Answer 1

您需要访问<meta name="csrf_token" content="{{ csrf_token() }}" /> $.ajaxSetup({ headers: { 'X-CSRF-TOKEN': $('meta[name="csrf_token"]').attr('content') } }); jQuery.ajax({ url: "{{route('check-email')}}", method: 'post', data: { username: jQuery('#username').val(), email: jQuery('#email').val() }, dataType: "json", success: function(result) { // Some Code } });属性，而不是尝试回显该对象。

public function checkEmail(Request $request)
{
    if($request->username) {
        $username = Members::where('member_nickname', $request->username)->count();
        $email = Members::where('member_email', $request->email)->count();

        if ($username > 0 && $email <=0) {
            return response()->json(['success' => 1]);
        }

        if ($username <= 0 && $email <=0) {
            return response()->json(['success' => 2]);
        }

        if ($email > 0 && $username <=0) {
            return response()->json(['success' => 3]);
        }

        if ($email > 0 && $username > 0) {
            return response()->json(['success' => 4]);
        }
    }
}

这将根据需要提供列名称。

阅读手册：https://php.net/manual/en/mysqli-result.fetch-field-direct.php

可恢复的致命错误：无法将类stdClass的对象转换为字符串

1 个答案: