请帮忙.我试图使用他们的学生ID显示每个用户的图像但它在我的页面上显示未知字符作为图像,如下所示
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$sql_up="SELECT content FROM sp_uploads WHERE s_id= 'RC49557'";
$result_set=mysql_query($sql_up);
while($row_up = mysql_fetch_array($result_set))
{
$pp = $row_up['content'];
}
echo "<td>
<img style=\"float:right; margin:5px; width:150px; height:150px; padding:10px; \"
src=\" data:image/jpg;charset=utf8;base64,<?php echo $pp ?> \" /> </td>";
Run Code Online (Sandbox Code Playgroud) 我想显示重复或不唯一的值的错误,但我的表单采用输入数组,我已经在jsfiddle检查了这些问题,name = "week[]"失败但name = "week"工作正常
表单HTML
<html>
<head></head>
<body>
<form name = "myForm" id = "myForm" class ="validate">
<input type="number" name="week[]" id="week1"/>
<input type="number" name="week[]" id="week2"/>
<input type="number" name="week[]" id="week3"/>
<input type="number" name="week[]" id="week4"/>
</form>
<script src="assets/js/jquery.validate.min.js"></script>
</body>
</html>
Run Code Online (Sandbox Code Playgroud)
我试过这个
<script type="text/javascript">
jQuery.validator.addMethod("unique", function(value, element, params) {
var prefix = params;
var selector = jQuery.validator.format("[name!='{0}'][unique='{1}']", element.name, prefix);
var matches = new Array();
$(selector).each(function(index, item) {
if (value == $(item).val()) {
matches.push(item);
} …Run Code Online (Sandbox Code Playgroud)