如何在php中显示从一个mysql表到多个mysql表的记录?
我在使用php从数据库中获取记录时遇到问题。但是,当我简单地在phpmyadmin sqls中执行以下查询时,它会工作,但当我尝试使用php执行时,它会出现如下错误: “未定义的索引体验。开始”如何在php中显示从一个mysql表到多个mysql表的记录?,php,mysql,Php,Mysql,我在使用php从数据库中获取记录时遇到问题。但是,当我简单地在phpmyadmin sqls中执行以下查询时,它会工作,但当我尝试使用php执行时,它会出现如下错误: “未定义的索引体验。开始” 您应该使用$row[“start”]instant of$row[“experience.start”]作为MYSQL返回列名,而不是表。列名 $experience = "SELECT experience.start, experience.finish, experience.isActive,
您应该使用$row[“start”]instant of$row[“experience.start”]作为MYSQL返回列名,而不是表。列名
$experience = "SELECT
experience.start,
experience.finish,
experience.isActive,
experience.title,
experience.nameOfOrgnization,
experienceprofessionnelle.organismeLogoURL,
description.details
FROM experience,description WHERE experience.ID_EP = description.ID_Description";
$result = $connection->query($experience);
while($row=$result->mysqli_fetch_array()){
echo $row["experience.start"];
}
或
$experience = "SELECT
experience.start,
experience.finish,
experience.isActive,
experience.title,
experience.nameOfOrgnization,
experienceprofessionnelle.organismeLogoURL,
description.details
FROM experience,description WHERE experience.ID_EP = description.ID_Description";
$result = $connection->query($experience);
while($row=$result->fetch_assoc()){
echo $row["start"];
}
在最后一行代码中,请尝试:
$row[“Date处女秀”]
。
$experience = "SELECT
experience.start,
experience.finish,
experience.isActive,
experience.title,
experience.nameOfOrgnization,
experienceprofessionnelle.organismeLogoURL,
description.details
FROM experience,description WHERE experience.ID_EP = description.ID_Description";
$result = $connection->query($experience);
while($row=$result->fetch_assoc()){
echo $row["start"];
}