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easyui-php如何获取数据库中blob，然后将他显示到datagrid中
- WBOYOriginal
- 2016-06-02 11:33:271179browse
easyuiphp上传图片datagrid
现在php已经将图片存入到数据库中，类型blob。
<code> <?php $sql = "select pic from e_user where uid = '1dff5b51f862e6d181577e3ca34248be'"; $js = get_js_object($sql); Header( "Content-type: image/png"); echo $js->pic; echo '<p><img src="../php/testlist.php" style="max-width:90%" alt="easyui-php如何获取数据库中blob，然后将他显示到datagrid中" ></p>'; ?></code>
<code> <table class="easyui-dategrid" url="../php/testlist.php"> <thead> <tr> <th field="pic" width="120">图片</th> </tr> </thead> </table></code>
如何修改一下？
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