Php 如何在无序列表中列出多个表结果
我试图从数据库中列出现有的用户配置文件,但只显示一个结果,而有多个记录Php 如何在无序列表中列出多个表结果,php,mysqli,Php,Mysqli,我试图从数据库中列出现有的用户配置文件,但只显示一个结果,而有多个记录 $SQL_Users = "SELECT DISTINCT first_party_code FROM t_planning"; $RESULT = mysqli_query( $conn, $SQL_Users); while($row_Users = mysqli_fetch_array($RESULT)) { $data0 = "&
$SQL_Users = "SELECT DISTINCT first_party_code FROM t_planning";
$RESULT = mysqli_query( $conn, $SQL_Users);
while($row_Users = mysqli_fetch_array($RESULT))
{
$data0 = "<ul class='list-unstyled users-list'>";
$SQL_User_Info = "SELECT title, first_name, middle_name, last_name, profile_image_name FROM t_employees WHERE employee_code = '" . $row_Users["first_party_code"] . "'";
$RESULT_Users = mysqli_query( $conn, $SQL_User_Info);
while($Users = mysqli_fetch_array($RESULT_Users))
{
$data0.= "<li data-toggle='tooltip' data-popup='tooltip-custom' data-placement='bottom' title='" . $Users["title"] . " " . $Users["first_name"] . " " . $Users["middle_name"] . " " . $Users["last_name"] . "' class='avatar pull-up'>
<img class='media-object rounded-circle' src='../uploads/users_profile_images/" . $Users["profile_image_name"] . "' alt='Avatar' height='30' width='30'>
</li>";
}
echo $data0.="</ul>";
mysqli_close($conn);
$SQL\u Users=“从计划中选择不同的第一方代码”;
$RESULT=mysqli\u查询($conn,$SQL\u用户);
while($row\u Users=mysqli\u fetch\u数组($RESULT))
{
$data0=“”;
$SQL_User_Info=“从t_员工中选择标题、名字、中间名字、姓氏、个人资料图像名称,其中员工代码=”。“$row_用户[“第一方代码”]。”;
$RESULT\u Users=mysqli\u query($conn,$SQL\u User\u Info);
而($Users=mysqli\u fetch\u数组($RESULT\u Users))
{
$data0.=“-
”;
}
echo$data0.=“
”;
mysqli_close($conn);
找到了问题所在。我过早地关闭了mysqli_close($conn)
$SQL_Users = "SELECT DISTINCT first_party_code FROM t_planning";
$RESULT = mysqli_query( $conn, $SQL_Users);
while($row_Users = mysqli_fetch_array($RESULT))
{
$data0 = "<ul class='list-unstyled users-list'>";
$SQL_User_Info = "SELECT title, first_name, middle_name, last_name, profile_image_name FROM t_employees WHERE employee_code = '" . $row_Users['first_party_code'] . "'";
$RESULT_Users = mysqli_query( $conn, $SQL_User_Info);
while($Users = mysqli_fetch_array($RESULT_Users))
{
$data0.= "<li data-toggle='tooltip' data-popup='tooltip-custom' data-placement='bottom' title='" . $Users["title"] . " " . $Users["first_name"] . " " . $Users["middle_name"] . " " . $Users["last_name"] . "' class='avatar pull-up'>
<img class='media-object rounded-circle' src='../uploads/users_profile_images/" . $Users["profile_image_name"] . "' alt='Avatar' height='30' width='30'>
</li>";
}
echo $data0.="</ul>";
}
mysqli_close($conn);
$SQL\u Users=“从计划中选择不同的第一方代码”;
$RESULT=mysqli\u查询($conn,$SQL\u用户);
while($row\u Users=mysqli\u fetch\u数组($RESULT))
{
$data0=“”;
$SQL_User_Info=“从t_员工中选择标题、名字、中间名字、姓氏、个人资料图像名称,其中员工代码=”。“$row_用户['first_party_code']。”;
$RESULT\u Users=mysqli\u query($conn,$SQL\u User\u Info);
而($Users=mysqli\u fetch\u数组($RESULT\u Users))
{
$data0.=“-
”;
}
echo$data0.=“
”;
}
mysqli_close($conn);
您正在使用变量$RESULT\u Users
两次——可能不是您想要的。即,后者正在覆盖前者。重命名了一个$RESULT\u Users,但仍然给出一个结果