Question

当试图从数据库中显示形象时,没有展示图像时,不知道问题是什么,即我的代码。

显示_desc.php

<?php 
  $errmsg = "";

  if (! @mysql_connect("localhost","root","")) {
    $errmsg = "Cannot connect to database";
  }

  @mysql_select_db("dbname");

  if (isset($_GET[ img_name ])) {
    $gotten = @mysql_query("select img from image where img_id = ".$_GET[ img_name ]);

    header("Content-type: image/x-ms-bmp");

    while ($row = mysql_fetch_array($gotten)) {
      print $row[ img ];
    }
    mysql_free_result($gotten);
  }
  ?>

显示.php

  <?php
    $errmsg = "";

    if (! @mysql_connect("localhost","root","")) {
      $errmsg = "Cannot connect to database";
    }

    @mysql_select_db("dbase_mgb");

    $strSQL = "select * from image";
    $rsPix = mysql_query($strSQL);
    $numRows = mysql_numrows($rsPix);

    $i = 0;

    while($i < $numRows) {
  ?>

  <image src="显示_desc.php?img_id=<?php echo mysql_result($rsPix,$i,"img_id"); ?>"

  <?php

    $i++;

    }
  ?>

谁能帮助我?

Answer 1

Your display.php doesn t seem to produce proper <img> tags.
Check that your img field in the database is marked as BINARY

Answer 2

i 您使用图像数据类型作为BLOB

Answer 3

为什么你们对数据库有强烈的图像? 最好是简单地将图像的名称/顺序保存在数据库中,将其上载到一个盒子上,并在检索<代码><img>标签时加以显示。

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