【发布时间】:2011-03-08 03:13:24
【问题描述】:
我仍在学习函数及其工作原理。我知道我做错了什么,只是不知道如何解决它。我正在编写一个函数来从数据库中提取图像数据并将其返回到屏幕上。它可以工作,但如果有多个图像,它将仅返回最后一个图像。我知道问题是$project_image 仅返回最后一张图像,因为 while 循环的工作方式,但我的问题是我如何才能不使用 while 循环或让它向$project_image 变量添加多个图像.
精简功能
function get_project_image($id,$type="thumb",$src="false",$limit=1){
if($type =="main"){
$project_image_qry = mysql_query(" SELECT i_name FROM `project_images` WHERE i_project_id = '$id' AND i_type= '2' LIMIT $limit " ) or die(mysql_error());
$project_image="";
while($project_image_row = mysql_fetch_array($project_image_qry)) {
$project_image_result = mysql_fetch_array ($project_image_qry);
if($src=="true"){
$project_image .= '<img src="'.admin_settings('site_url').admin_settings('image_main_dir').'/'.$project_image_result['i_name'].'" alt="project_image"/>';
}
else{
$project_image .= $project_image_result['i_name'];
}
}
}
return $project_image;
}
【问题讨论】:
标签: php function while-loop