SQL查询在PHP中返回false [英] SQL query returning false in PHP
问题描述
我正在尝试在PHP中执行此查询，但是它始终返回false.我已经在phpMyAdmin中尝试了该查询，它的工作原理很好，因此，如果有人可以发现错误，那将是很好的.另外，如何获得类似此类问题的更好的错误消息，以便我可以尝试解决问题?
I am trying to perform this query in PHP however it keeps returning false. I have tried the query in phpMyAdmin and it works fine so if anyone can spot what is wrong that would be great. Also how can I get some better error messages for problems like this so I can try and solve the problem?
$stmt = $conn->prepare("SELECT * FROM artist WHERE artID != ?;");
if(!$stmt)
{
echo "Error creating SQL statement";
return 1;
}
我已经在同一段PHP代码中使用$stmt = $conn->prepare(query);进行了不同的查询，该查询运行良好，所以我不知道这是否与它有关.
I have already used $stmt = $conn->prepare(query); for a different query in the same block of PHP code which runs fine so I don't know if that is anything to do with it.
先谢谢您了:)
有人问我在哪里绑定?"在查询中使用. $stmt->bind_param('i', $albumArtID);我之所以没有在问题中包括它，是因为if语句中的echo运行，所以我认为它在bind_param之前遇到了错误.
I was asked where I bind the '?' used in the query. $stmt->bind_param('i', $albumArtID); I didn't include it in the question originally because the echo in the if statement runs so I presumed it was encountering an error before the bind_param.
以下是用于建立连接的代码:
EDIT 2: As requested here is the code used to make the connection:
<?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>";
?>
这是该页面中代码的整个主要部分，希望我们能弄清楚这一点:
EDIT 3: Here is the entire main section of code from that page, hopefully we can figure this out:
<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);
$stmt->execute();
}
?>
</form>
推荐答案
在第一个查询完成后，使用mysqli_close($conn);关闭第一个连接，然后在第二个查询之前使用include 'connection.php';打开一个新连接.归功于@ Chay22
Close first connection using mysqli_close($conn); after first query is finished then open a new connection with include 'connection.php'; before the second query. Credit to @Chay22
这篇关于SQL查询在PHP中返回false的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
