我试图在PHP中执行此查询，但它一直返回false。我已经尝试了phpMyAdmin中的查询，它运行正常，所以如果有人能发现什么是错的那将是伟大的。另外，我如何才能获得更好的错误信息，以便解决问题？
$stmt = $conn->prepare("SELECT * FROM artist WHERE artID != ?;");
if(!$stmt)
{
echo "Error creating SQL statement";
return 1;
}
我已经在同一个PHP代码块中使用$stmt = $conn->prepare(query);进行了不同的查询，运行正常，所以我不知道这是否与它有关。
提前致谢：）
编辑：有人问我在哪里绑定了＆＃39;？＆＃39;在查询中使用。$stmt->bind_param('i', $albumArtID);我最初没有在问题中包含它，因为echo语句中的if运行所以我假设它在bind_param之前遇到错误。< / p>
编辑2：这里要求的是用于建立连接的代码：
<?php
error_reporting(-1);
ini_set('display_errors', 'On');
$db_host = 'localhost';
$db_user = 'root';
$db_pass = '';
$db_name = 'psyjb6';
$conn = new mysqli('localhost', 'root', '', 'psyjb6');
if ($conn->connect_errno)
echo"<p>failed to connect to database</p>";
?>
编辑3：这是该页面代码的完整主要部分，希望我们可以解决这个问题：
<form name="editAlbum" method="get" onsubmit="return validateForm(this)">
<div class="row">
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
include 'connection.php';
if(isset($_GET["album"]))
{
/* If album was passed in the URL then get current values
for that album */
$stmt = $conn->prepare("SELECT cd.artID, artName, cdTitle, cdPrice, cdGenre, cdTracks FROM cd INNER JOIN artist ON (cd.artID = artist.artID AND cdID = ?);");
if(!$stmt)
{
echo "Error creating SQL statement";
exit;
}
$albumID = htmlspecialchars($_GET["album"]);
$stmt->bind_param('i', $albumID);
$stmt->execute();
$stmt->bind_result($albumArtID, $albumArtName, $albumTitle,
$albumPrice, $albumGenre, $numTracks);
$stmt->fetch();
/* Create input fields */
// Album Title
echo "<div class=\"row horizontal-center\">" .
"<input type=\"text\" value=\"" . htmlspecialchars($albumTitle) . "\" name=\"albumTitle\"/>" .
"</div>";
// Artist Name
echo "<div class=\"row horizontal-center\">" .
"<h6>By Artist:</h6>" .
"</div>";
echo "<div class=\"row horizontal-center\">" .
"<select name=\"artID\">";
/* Create option for current artist so it will be first in list */
echo "<option value=\"$albumArtID\">$albumArtName</option>\n";
/* Generate list of artists except artist currently associated with the album */
$stmt = $conn->prepare("SELECT * FROM artist WHERE artID != ?");
if($stmt === false)
{
echo $conn->error;
echo "hi";
exit;
}
$stmt->bind_param('i', $albumArtID);
$stmt->execute();
$stmt->bind_result($artID, $artName);
/* Check if no artists were found */
if(!$stmt->fetch())
echo "<p>No artists were found!</p>";
else
{
/* Create options for artists that were found */
do
{
echo "<option value=\"$artID\">$artName</option>\n";
}while($stmt->fetch());
}
echo "</select>" .
"</div>";
// Album Price
echo "<div class=\"row horizontal-center\">" .
"<input type=\"number\" step=\"0.01\" value=\"" . htmlspecialchars($albumPrice) . "\" name=\"albumPrice\"/>" .
"</div>";
// Album Genre
echo "<div class=\"row horizontal-center\">" .
"<input type=\"text\" value=\"" . htmlspecialchars($albumGenre) . "\" name=\"albumGenre\"/>" .
"</div>";
// Number of Tracks
echo "<div class=\"row horizontal-center\">" .
"<input type=\"number\" value=\"" . htmlspecialchars($numTracks) . "\" name=\"numTracks\"\n/>" .
"</div>";
// Delete checkbox
echo "<div class=\"row\">" .
"<div class=\"col-2\">" .
"<h6>Delete:</h6>" .
"</div>" .
"<div class=\"col-1\">" .
"<input type=\"checkbox\" name=\"delete\" value=\"Delete\"/>" .
"</div>" .
"</div>";
/* Create hidden field to submit the album ID with the form */
echo "<input type=\"hidden\" value=\"" . htmlspecialchars($albumID) . "\" name=\"albumID\"\n/>";
}
else
{
/* Send browser back to artists page if they somehow accessed
the edit page without going through the "Edit" link next
to an artist in the table. This would be the artName variable
would not be sent via the URL.*/
header("Location: artists.php");
}
?>
</div>
<div class="row">
<div class="col-2">
<h6>Delete:</h6>
</div>
<div class="col-1">
<input type="checkbox" name="delete" value="Delete"/>
</div>
</div>
<div class="row">
<input type="submit" name="submit" value="Update"/>
</div>
<!-- PHP to edit album data -->
<?php
include 'connection.php';
if(isset($_GET["delete"]))
{
$albumID = $_GET["albumID"];
/* Create DELETE query */
$stmt = $conn->prepare("DELETE FROM cd WHERE cdID = ?;");
if(!$stmt)
{
echo "Error creating SQL statement";
exit;
}
$stmt->bind_param('i', $albumID);
$stmt->execute();
}
else if(isset($_GET["albumTitle"]) && isset($_GET["albumGenre"])
&& isset($_GET["albumPrice"]) && isset($_GET["numTracks"]))
{
$albumTitle = htmlspecialchars($_GET["albumTitle"]);
$artID = htmlspecialchars($_GET["artID"]);
$albumGenre = htmlspecialchars($_GET["albumGenre"]);
$albumPrice = htmlspecialchars($_GET["albumPrice"]);
$numTracks = htmlspecialchars($_GET["numTracks"]);
/* Create INSERT query */
$stmt = $conn->prepare("UPDATE cd SET (cdTitle = ?, artID = ?,
cdGenre = ?, cdPrice = ?, cdTracks = ?) WHERE cdID = ?;");
if(!$stmt)
{
echo "Error creating SQL statement";
exit;
}
$stmt->bind_param('sisdi', $albumTitle, $artID, $albumGenre,
$albumPrice, $numTracks);
$stmt->execute();
}
?>
</form>
答案 0 :(得分：0)
如果使用参数化查询，则必须在执行准备好的查询时传递参数的值。
您还必须execute准备好的查询。 prepare只是将查询传递给数据库进行编译和优化，它实际上并不执行查询。
此外，如果您在这些数据库访问语句中出现错误，则应使用函数/方法来显示确实错误消息，这些消息比输出您自己编写的内容（echo "Error creating SQL statement"; <更有用。 / p>
此外，;也不是必需的。
$stmt = $conn->prepare("SELECT * FROM artist WHERE artID != ?");
if ( $stmt === false ){
echo $conn->error;
exit;
}
$stmt->bindParam('i', $some_variable)
$result = $stmt->execute();
if ( $result === false ) {
echo $stmt->error;
exit;
}
答案 1 :(得分：0)
在第一次查询完成后，使用mysqli_close($conn);关闭第一个连接，然后在第二个查询之前打开与include 'connection.php';的新连接。感谢@ Chay22