php - 如何更改jpg质量导出？

Question

在下面的代码中，将 jpg 存储在 db 中，我需要知道如何先将 jpg 更改为 40% 质量然后保存？我试过 imagejpg ，但它保存空文件:

function exportGraphics($table_name)
{
    $odbc_query = "SELECT * FROM " . $table_name;
    mkdir("TI/" . $table_name);

    $data = odbc_exec($this->odbc_id, $odbc_query);
    odbc_longreadlen($data, 10485760); // 10MB = 10485760
    while (odbc_fetch_row($data)) {
        $row = odbc_fetch_array($data);
        if ($row['GRD_ID'] != "") {
            $file_name_jpg = "TI/" . $table_name . "/" . $row['GRD_ID'] . ".jpg";
            $file = fopen($file_name_jpg, "w");
            fputs($file, $row['GRD_GRAPHIC']);
            fclose($file);
            set_time_limit(3600);
            unset($row);
        }
    }
    print "Ýêñïîðò êàðòèíîê èç òàáëèöû " . $table_name . " çàâåðøåí!";
}

Warning: imagecreatefromstring() [function.imagecreatefromstring]: gd warning: one parameter to a memory allocation multiplication is negative or zero, failing operation gracefully in X:\denwer\www\denwer\tecdoc3.php on line 103

Warning: imagecreatefromstring() [function.imagecreatefromstring]: Passed data is not in 'WBMP' format in X:\denwer\www\denwer\tecdoc3.php on line 103

Warning: imagecreatefromstring() [function.imagecreatefromstring]: Couldn't create GD Image Stream out of Data in X:\denwer\www\denwer\tecdoc3.php on line 103

Warning: imagecreatefromstring() [function.imagecreatefromstring]: gd warning: one parameter to a memory allocation multiplication is negative or zero, failing operation gracefully in X:\denwer\www\denwer\tecdoc3.php on line 103

原代码:

 function exportGraphics($table_name) {
 $odbc_query = "SELECT * FROM " . $table_name;
 mkdir("TI/" . $table_name);

 $data = odbc_exec($this->odbc_id, $odbc_query);
 odbc_longreadlen($data, 10485760); //10MB = 10485760
 while(odbc_fetch_row($data)) 
 { 
 $row = odbc_fetch_array($data);
 if($row['GRD_ID'] != "") {
 $file_name_jp2 = "TI/" . $table_name . "/" . $row['GRD_ID'] . ".jp2";
 $file = fopen ($file_name_jp2, "w");
 fputs($file, $row['GRD_GRAPHIC']);
 fclose($file);
 set_time_limit(0);
 unset($row);
 }
 }
 print "Ýêñïîðò êàðòèíîê èç òàáëèöû " . $table_name . " çàâåðøåí!";
 }

Answer 1

试试这个:

function exportGraphics($table_name) {
    $odbc_query = "SELECT * FROM " . $table_name;
    mkdir("TI/" . $table_name);

    $data = odbc_exec($this->odbc_id, $odbc_query);
    odbc_longreadlen($data, 10485760); //10MB = 10485760
    while(odbc_fetch_row($data))
    {
        $row = odbc_fetch_array($data);
        if($row['GRD_ID'] != "") {
            $file_name_jpg = "TI/" . $table_name . "/" . $row['GRD_ID'] . ".jpg";

            // create GD graphic from string, call imagejpeg to save new image
            $im = imagecreatefromstring($row['GRD_GRAPHIC']);
            imagejpeg($im, $file_name_jpg, 40);

            set_time_limit(3600);
            unset($row);
        }
    }
    print "Ýêñïîðò êàðòèíîê èç òàáëèöû " . $table_name . " çàâåðøåí!";
}

或者这正是您尝试过的？

如果由于某种原因不起作用，您可以尝试:

$im = imagecreatefromstring($row['GRD_GRAPHIC']);
ob_start();
imagejpeg($im, null, 40);
$imgData = ob_get_contents();
ob_end_clean();

$file = fopen ($file_name_jpg, "w+b");
fputs($file, $imgData);
fclose($file);

编辑:GD 不工作的原因是因为它不支持 JPEG-2000 文件格式

可能的解决方案:在服务器上安装 imagemagick 并尝试这样的代码:

$file_name_jpg = "TI/" . $table_name . "/" . $row['GRD_ID'] . ".jp2";
$file_out_jpg  = str_replace('.jp2', '.jpg', $file_name_jpg);

file_put_contents($file_name_jpg, $row['GRD_GRAPHIC']);

// execute imagemagick convert to change to jpeg with quality 40
exec("/usr/bin/convert $file_name_jpg -quality 40 -format jpg $file_out_jpg");
unlink($file_name_jpg); // get rid of temp jp2 file