[PHP + MySQL]数据库SELECT查询未返回结果 [英] [PHP + MySQL]Database SELECT query returning no result

问题描述

我正在尝试访问我的数据库以获取一些数据，但是它不断返回以下错误.

I am trying to access my database to get some data, but it keeps returning with the following errors.

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\default.php on line 84

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\default.php on line 86

我已经检查了连接，并且代码正在正确输入数据，只是结果查询不会返回任何值.就个人而言，我看不到错误在哪里，因为其他查询(例如INSERT和CREATE)运行良好.

I have checked the connection and the code is inputting data properly, it's just the results query that won't return any values. Personally, I can't see where the error is because other queries, such as INSERT and CREATE are working perfectly.

<?php
    mysqli_select_db($conn, $dbName);
    $sql = "SELECT * FROM tbl_users WHERE id = 1;";
    $result = mysqli_query($conn, $sql);
    echo mysqli_num_rows($result); //Line 84

    if (mysqli_num_rows($result) > 0) { //Line 86
        while($row = mysqli_fetch_assoc($result)) {
            ...
        }
    } else {
        echo "0 results";
    }
?>

如果您需要任何其他信息，请询问我，我将尝试提供.

If you require any further information, please ask me and I will attempt to provide it.

完整代码:

//**Create Connection**//
$conn = mysqli_connect($serverName, $username, $password);

//**Check Connection**//
if (!$conn) { die("Connection failed: " . mysqli_connect_error()); }
else { echo "<p>Connected successfully!</p>"; }



//**Create Database**//
$dbName = "myDB";
$sql = "CREATE DATABASE IF NOT EXISTS " . $dbName . " CHARACTER SET utf8 COLLATE utf8_general_ci;";

//Error Handling
if (!mysqli_query($conn, $sql)) { echo "Error creating database: " . mysqli_error($conn); }
else { echo "<p>Database created successfully!</p>"; }

//**Create Table**//
mysqli_select_db($conn, $dbName);
$tbl_name = "tbl_users";
$sql = "CREATE TABLE IF NOT EXISTS " . $tbl_name . " (id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, firstName VARCHAR(64) NOT NULL, lastName VARCHAR(64) NOT NULL, userEmail VARCHAR(256) NOT NULL, reg_date TIMESTAMP DEFAULT CURRENT_TIMESTAMP) CHARACTER SET utf8 COLLATE utf8_general_ci;";

//Error Handling
if (!mysqli_query($conn, $sql)) { echo "Error creating table: " . mysqli_error($conn); }
else { echo "<p>Table '" . $tbl_name . "' created successfully!</p>"; }

mysqli_select_db($conn, $dbName);
$sql = "SELECT * FROM tbl_users WHERE id = 1";
$result = mysqli_query($conn, $sql) or die(mysqli_error($conn));

if (!$result = mysqli_query($conn, $sql)) {
printf("Errormessage: %s\n", mysqli_error($conn));
}

if (mysqli_num_rows($result) > 0) {
    while($row = mysqli_fetch_assoc($result)) {
        echo $row . "<br>";
    }
} else {
    echo "0 results";
}

推荐答案

问题在这里:

您在这段代码中两次使用了mysqli_query():

You're using mysqli_query() twice in this piece of code:

$result = mysqli_query($conn, $sql) or die(mysqli_error($conn));

if (!$result = mysqli_query($conn, $sql)) {
printf("Errormessage: %s\n", mysqli_error($conn));
}

您需要删除一个并执行以下操作:

You need to remove one and do:

if (!$result) {
printf("Errormessage: %s\n", mysqli_error($conn));
}

同时添加逃生路线else{...}和 affected_rows() 也.

while adding an escape route else{...} and affected_rows() also.

• 这是查询失败的原因.

这是您使用$conn两次，而不是在查询中使用变量引用:

This, you're using $conn twice and not using a variable reference for your query:

mysqli_select_db($conn, $dbName);
$sql = "SELECT * FROM tbl_users WHERE id = 1";
$result = mysqli_query($conn, $sql) or die(mysqli_error($conn));

将其更改为并使用 mysqli_connect() ，因为此时已经创建了数据库:(假设ID上面已经创建了ID"1").

Change it to and using the 4 parameters scheme of mysqli_connect(), since the DB has already been created at this point: (assuming the id of "1" has already been created above that).

$dbName = "myDB";
$conn = mysqli_connect($serverName, $username, $password, $dbName);

$sql = "SELECT * FROM tbl_users WHERE id = 1";
$result = mysqli_query($conn, $sql) or die(mysqli_error($conn));

if (!$result) {
printf("Errormessage: %s\n", mysqli_error($conn));
}
else{ echo "Success"; }

或删除or die(mysqli_error($conn))并让错误继续传递(如果有的话).

or by removing or die(mysqli_error($conn)) and getting the error passed on after, if any.

$dbName = "myDB";
$conn = mysqli_connect($serverName, $username, $password, $dbName);

$sql = "SELECT * FROM tbl_users WHERE id = 1";
$result = mysqli_query($conn, $sql);

if (!$result) {
printf("Errormessage: %s\n", mysqli_error($conn));
}
else{ echo "Success"; }

(其他修改)

您也可以尝试此方法，并将其与我上面已经说过的结合使用:

You could also try this method and used in conjunction with what I already stated above:

$numrows = mysqli_num_rows($result);

if($numrows > 0){
// do something
}

---

将 错误报告 添加到文件顶部，这将有助于发现错误.

---

Add error reporting to the top of your file(s) which will help find errors.

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

侧注:错误报告仅应在登台进行，而绝不能在生产中进行.

Sidenote: Error reporting should only be done in staging, and never production.

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