我正在字符串中获取外键,因为我需要将其用于库,在库中,每个上传的图像都分配给一个摄影师,这些摄影师也需要显示在菜单中。
我遵循了这个指南,一切都很顺利。我现在需要用PHP输出数据——这我想不通。
<?php
$sql = "SELECT * FROM borrowed WHERE employee.id = 'Reck' JOIN employee ON employee.id = borrowed.employeeid";
$result = mysqli_query($db, $sql);
while($row = mysqli_fetch_array($result)) {
?>
<? echo $row['lastname']; ?>
<?php
}
?>
我得到错误警告:mysqli_fetch_array()要求参数1为mysqli_result,在/Applications/MAMP/htdocs/galleri/test.php的第12行给出布尔值
第12行是while循环。
查询中有错误,请参阅查询结构如何使用联接
在查询中,join
在where
条件之后,但必须在where
之前。它应该像这个
$sql = "SELECT * FROM borrowed JOIN employee ON employee.id = borrowed.employeeid WHERE employee.id = 'Reck'";
和你的代码
<?php
$sql = "SELECT * FROM borrowed JOIN employee ON employee.id = borrowed.employeeid WHERE employee.id = 'Reck'";
$result = mysqli_query($db, $sql);
while($row = mysqli_fetch_array($result)) {
echo $row['lastname'];
}
?>