要从1个表中选择数据,我使用以下代码:
if(isset($UserID)) {
$users = $con->prepare("
SELECT DISTINCT
d.FirstName
,d.LastName
,d.Picture
FROM Details d
WHERE d.UserId = ?
");
$users->bind_param('i', $GetUserId);
$users->execute();
$users->bind_result(
$FirstName,
$LastName,
$Picture
);
} else {
echo "There is no User ID detected, try to refresh browser.";
}
while ($users->fetch()) { ?>
<div class="grid-group">
<span>Personal Details</span>
<div class="grid-column">
<div class="grid-item header">First Name </div>
<div class="grid-item header">Last Name </div>
<div class="grid-item header">Picture </div>
</div>
<div class="grid-column">
<div class="grid-item"><?php echo $FirstName; ?></div>
<div class="grid-item"><?php echo $LastName; ?></div>
<div class="grid-item"><?php echo $Picture; ?></div>
</div>
</div>
<?php } ?>
如何再添加一个查询以从另一个表中选择一些数据?类似于:
SELECT Foo, Bar, FooBar
FROM Bizz
WHERE UserId = ?
我应该使用mysqli_multi_query
吗?但是,我如何使用它和它之后的bind_result
?我无法实现正确的语法。或者还有其他办法?
我需要打印两个表中的数据,如下所示:
while ($users->fetch()) { ?>
<div class="grid-group">
<span>Personal Details</span>
<div class="grid-column">
<div class="grid-item header">First Name </div>
<div class="grid-item header">Last Name </div>
<div class="grid-item header">Picture </div>
</div>
<div class="grid-column">
<div class="grid-item"><?php echo $FirstName; ?></div>
<div class="grid-item"><?php echo $LastName; ?></div>
<div class="grid-item"><?php echo $Picture; ?></div>
</div>
</div>
<div class="grid-group">
<span>Foo Bar</span>
<div class="grid-column">
<div class="grid-item header">Foo </div>
<div class="grid-item header">Bar </div>
<div class="grid-item header">FooBar </div>
</div>
<div class="grid-column">
<div class="grid-item"><?php echo $Foo; ?></div>
<div class="grid-item"><?php echo $Bar; ?></div>
<div class="grid-item"><?php echo $FooBar; ?></div>
</div>
</div>
<?php } ?>
由于某些原因,我不能使用JOINS
。你有什么想法吗?
更新
如果我在下面准备了两个查询,我会得到一个空白屏幕:
if(isset($UserID)) {
$users = $con->prepare("
SELECT DISTINCT
d.FirstName
,d.LastName
,d.Picture
FROM Details
WHERE d.UserId = ?
");
$users->bind_param('i', $GetUserId);
$users->execute();
$users->bind_result(
$FirstName,
$LastName,
$Picture
);
$users2 = $con->prepare("
SELECT DISTINCT
Foo
, Bar
, FooBar
FROM Bizz
WHERE UserId = ?
");
$users2->bind_param('i', $GetUserId);
$users2->execute();
$users2->bind_result(
$Foo,
$Bar,
$FooBar
);
} else {
echo "There is no User ID detected, try to refresh browser.";
}
while ($users->fetch()) { ?>
...
你有什么想法吗?
要知道为什么会出现空白屏幕,必须打开error_reporting,请在php代码的开头添加以下内容,然后再添加其他内容:
error_reporting(-1);
还添加一些代码来显示MySQL中可能发生的错误:
if(isset($UserID)) {
$users = $con->prepare("
SELECT DISTINCT
d.FirstName
,d.LastName
,d.Picture
FROM Details
WHERE d.UserId = ?
");
if (!$users) {
echo 'MySQL Connect Error in Query: (' . $mysqli->errno . ') ';
}
$binding = $users->bind_param('i', $GetUserId);
if (!$binding) {
echo 'MySQL Connect Error in Query: (' . $mysqli->errno . ') ';
}
if($users->execute() == false) {
echo 'Error: ' . $con->error;
}
$binding_results = $users->bind_result(
$FirstName,
$LastName,
$Picture
);
if (!$binding_results) {
echo 'MySQL Connect Error in Query: (' . $mysqli->errno . ') ';
}
$users2 = $con->prepare("
SELECT DISTINCT
Foo
, Bar
, FooBar
FROM Bizz
WHERE UserId = ?
");
if (!$users2) {
echo 'MySQL Connect Error in Query: (' . $mysqli->errno . ') ';
}
$binding2 = $users2->bind_param('i', $GetUserId);
if (!$binding2) {
echo 'MySQL Connect Error in Query: (' . $mysqli->errno . ') ';
}
if($users2->execute() == false) {
echo 'Error: ' . $con->error;
}
$binding_results2 = $users2->bind_result(
$Foo,
$Bar,
$FooBar
);
if (!$binding_results2) {
echo 'MySQL Connect Error in Query: (' . $mysqli->errno . ') ';
}
我可以在你的第一个问题中看到一个错误。您使用的是d.FirstName,其中表名为Details