警告:mysql_fetch_array()期望参数1是资源... 当我试图从数据库获取下面的第一个特色配置文件的值时,会显示以上错误..... 如果我添加: - 或死(mysql_error()); 然后它显示没有选择数据库.. 请帮忙 代码: -
<div class="grid_1">
<div class="container">
<h1>Featured Profiles</h1>
<div class="heart-divider">
<span class="grey-line"></span>
<i class="fa fa-heart pink-heart"></i>
<i class="fa fa-heart grey-heart"></i>
<span class="grey-line"></span>
</div>
<?php
$sql="SELECT (file,user_name,user_id,user_occupation) FROM users1";
$result_set=mysql_query($sql) or die(mysql_error());
while($row=mysql_fetch_object($result_set))
{
?>
<ul id="flexiselDemo3">
<li><div class="col_1"><a href="view_profile.html">
<img src="images/<?php echo $row->file ?>" alt="" class="hover-animation image-zoom-in img-responsive"/>
<div class="layer m_1 hidden-link hover-animation delay1 fade-in">
<div class="center-middle">About Him</div>
</div>
<h3><span class="m_3">Profile ID : <?php echo $row->user_id ?></span><br>28, Christian, Australia<br>Corporate</h3></a></div>
</li>
<li><div class="col_1"><a href="view_profile.html">
<img src="images/2.jpg" alt="" class="hover-animation image-zoom-in img-responsive"/>
<div class="layer m_1 hidden-link hover-animation delay1 fade-in">
<div class="center-middle">About Her</div>
</div>
<h3><span class="m_3">Profile ID : MI-387412</span><br>28, Christian, Australia<br>Corporate</h3></a></div>
</li>
<li><div class="col_1"><a href="view_profile.html">
<img src="images/3.jpg" alt="" class="hover-animation image-zoom-in img-responsive"/>
<div class="layer m_1 hidden-link hover-animation delay1 fade-in">
<div class="center-middle">About Him</div>
</div>
<h3><span class="m_3">Profile ID : MI-387412</span><br>28, Christian, Australia<br>Corporate</h3></a></div>
</li>
<li><div class="col_1"><a href="view_profile.html">
<img src="images/4.jpg" alt="" class="hover-animation image-zoom-in img-responsive"/>
<div class="layer m_1 hidden-link hover-animation delay1 fade-in">
<div class="center-middle">About Her</div>
</div>
<h3><span class="m_3">Profile ID : MI-387412</span><br>28, Christian, Australia<br>Corporate</h3></a></div>
</li>
<li><div class="col_1"><a href="view_profile.html">
<img src="images/5.jpg" alt="" class="hover-animation image-zoom-in img-responsive"/>
<div class="layer m_1 hidden-link hover-animation delay1 fade-in">
<div class="center-middle">About Him</div>
</div>
<h3><span class="m_3">Profile ID : MI-387412</span><br>28, Christian, Australia<br>Corporate</h3></a></div>
</li>
<li><div class="col_1"><a href="view_profile.html">
<img src="images/6.jpg" alt="" class="hover-animation image-zoom-in img-responsive"/>
<div class="layer m_1 hidden-link hover-animation delay1 fade-in">
<div class="center-middle">About Her</div>
</div>
<h3><span class="m_3">Profile ID : MI-387412</span><br>28, Christian, Australia<br>Corporate</h3></a></div>
</li>
</ul>
<script type="text/javascript">
$(window).load(function() {
$("#flexiselDemo3").flexisel({
visibleItems: 6,
animationSpeed: 1000,
autoPlay:false,
autoPlaySpeed: 3000,
pauseOnHover: true,
enableResponsiveBreakpoints: true,
responsiveBreakpoints: {
portrait: {
changePoint:480,
visibleItems: 1
},
landscape: {
changePoint:640,
visibleItems: 2
},
tablet: {
changePoint:768,
visibleItems: 3
}
}
});
});
</script>
<script type="text/javascript" src="js/jquery.flexisel.js"></script>
<?php
}
?>
</div>
</div>
答案 0 :(得分:1)
我认为问题出在你的$ result函数中它应该有2个参数而你只给了一个所以你的代码应该是这样的
<?php
$connection=mysqli_connect('localhost','yourmysqlusername','yourmysqlpassword','yourmysqldatabase');
$sql="SELECT (file,user_name,user_id,user_occupation) FROM users1";
//you have to add $connection as 2nd parameter in $result function
$result_set=mysql_query($connection,$sql) or die(mysql_error());
while($row=mysql_fetch_object($result_set))
?>
答案 1 :(得分:0)
确保首先使用mysql_select
连接到数据库,然后使用mysql_select_db
选择数据库。在此之后,您需要使用mysql_query
启动查询。
此外,您可以mysqli_connect
组合前两个步骤。见this