从表联接获取信息以在PHP中显示时出现问题
{将两个表连接在一起(带教师ID和教师姓名) 我已经创建了连接,并在SQL中进行了测试。看起来不错。我正在尝试在屏幕上打印它从表联接获取信息以在PHP中显示时出现问题,php,mysql,Php,Mysql,{将两个表连接在一起(带教师ID和教师姓名) 我已经创建了连接,并在SQL中进行了测试。看起来不错。我正在尝试在屏幕上打印它 $classandteacher = "SELECT person_name FROM people RIGHT OUTER JOIN classes ON classes.instructor_id=people.instructor_id ASC"; $result = mysqli_query($dbc, $classandteacher){
$classandteacher = "SELECT person_name FROM people RIGHT OUTER JOIN classes ON classes.instructor_id=people.instructor_id ASC";
$result = mysqli_query($dbc, $classandteacher){
while($row = mysqli_query($dbc, $result));
$teacher = $row["person_name"];
echo ("Teacher: " . $teacher . "<br>");}
$classandteacher=“从人员右侧外部选择人员\u名称加入课程。讲师\u id=people.讲师\u id ASC”;
$result=mysqli\u查询($dbc,$classandteacher){
while($row=mysqli_query($dbc,$result));
$teacher=$row[“人名”];
回声(“教师:“.$Teacher.”
”;)
这是因为While循环的格式不正确
移除;
并将其放入{}
您还需要使用mysqli\u fetch\u assoc($result)
从查询中获取数组
$classandteacher = "SELECT person_name FROM people RIGHT OUTER JOIN classes ON classes.instructor_id=people.instructor_id ASC";
$result = mysqli_query($dbc, $classandteacher);
if(!$result) { //If your MYSQL query is throwing an error
error_log(mysqli_error());
}
while($row = mysqli_fetch_assoc($result))
{
$teacher = $row["person_name"];
echo "Teacher: " . $teacher . "<br>";
}
$classandteacher=“从人员右侧外部选择人员\u名称加入课程。讲师\u id=people.讲师\u id ASC”;
$result=mysqli_查询($dbc,$classandteacher);
if(!$result){//如果您的MYSQL查询抛出错误
错误日志(mysqli_error());
}
while($row=mysqli\u fetch\u assoc($result))
{
$teacher=$row[“人名”];
呼应“教师:.”教师“
”;
}
或尝试使用
改为mysqli_fetch_数组
$classandteacher =
"SELECT
person_name FROM
people RIGHT OUTER JOIN classes ON
classes.instructor_id
=
peopleinstructor_id ASC";
$result =
mysqli_query($dbc,$ classandteacher);
if(!$result) { //If
your MYSQL query is
throwing an error
error_log(mysqli_er
ror() );}
while($row =
mysqli_fetch_array(
$result))
{
$teacher = $row
["person_name"];
echo "Teacher: " . $
teacher . "<br>"; }
$classandteacher=
“选择
人名
右边的人在外面上课
班级.讲师id
=
人民教官(身份证)";
$result=
mysqli_查询($dbc,$classandteacher);
if(!$result){//if
您的MYSQL查询是
抛出错误
错误日志(mysqli)
ror());}
而($行=
mysqli_fetch_数组(
$结果)
{
$teacher=$row
[“人名”];
回声“老师:”$
教师。“
”;}
ooh…;
之后,外部联接一个表(从中不选择任何列)是没有意义的输出错误?您可能需要检查$result
是否返回FALSE
,然后使用mysqli\u error()
若要查看您的查询是否有问题,请参阅更新的回答无错误。连接良好。select语句良好。抱歉,如果没有任何错误,将无法提供更多帮助!请确保您的错误报告已打开。错误报告(1)
感谢您的帮助。