mysqli_query無法識別數據庫？

Question

我已經就此代碼提出了一個問題，我正在處理，而不是針對同一問題。無論哪種方式抱歉轉發！

所以我在與代碼的麻煩，具體如下：

<?php 
// Create connection 

$host = "localhost"; 
$username="tudor"; 
$password="passw0rd"; 

$con=mysqli_connect($host, $username, $password); 
if(! $con) 
{ 
    die('Could not connect: ' . mysqli_error()); 
} 
echo 'Connected successfully<br />'; 


$db_1 = mysqli_select_db($con, 'db_1'); 
if (! $db_1) { 
die('Could not select database: ' . mysqli_error()); 
} 
else { 
echo "Database successfully selected<br />===============================<br />"; 
} 

//=================================== 



$a = 1; 
$b = 2234; 

$table = "CREATE TABLE info (id INT NOT NULL AUTO_INCREMENT, city CHAR(40), country CHAR(40))"; 
if (! $table) { 
die('Could not create table ' . mysqli_error($con)); 
} 
else { 
echo "Table created<br />"; 
} 

$insert = "INSERT INTO info (city, country) VALUES ($a, $b)"; 
if (! $insert) { 
die('Could not insert ' . mysqli_error($con)); 
} 
else { 
echo "Inserted<br />"; 
} 

$select = "SELECT * FROM info"; 


$result = mysqli_query ($con, $insert); 
if (! $result) { 
die('Result not working ' . mysqli_error($con)); 
} 
else { 
echo "Result working<br />"; 
} 

echo "result: ".$result['city']. " "; 


mysqli_close($con); 

?> 

這個輸出（塊引用不顯示分頁符）：

連接成功數據庫成功入選 == =============================表創建插入結果不工作表'db_1.info'不存在

什麼d oes它的意思是「表'db.info'」不存在？它清楚地表明我的信息表已創建... 我試圖做的是顛倒$ result查詢中的變量：$ result = mysqli_query（$ insert，$ con）;因爲我曾在一本書中看過這種語法。然而，所有這給在輸出以下消息：

警告：mysqli_query（）預計參數1是mysqli的，給予串用C ：\ WAMP ...

思考的人？提前致謝！

編輯：非常感謝大家的幫助，非常感謝！

Answer 1

確定以獲取從查詢數組。這是你的代碼。

if(! $con) 
{ 
    die('Could not connect: ' . mysqli_error()); 
} 
echo 'Connected successfully<br />'; 
if (!$con) {trigger_error("Could not connect to MySQL: " . mysqli_connect_error()); } 
else { echo "Database successfully connected<br />===============================<br />"; } 
$a = 1; 
$b = 2234; 
$table = mysqli_query($con,"CREATE TABLE IF NOT EXISTS info (`id` int(11) unsigned NOT NULL auto_increment, 
`city` CHAR(40), 
`country` CHAR(40), PRIMARY KEY (`id`))ENGINE=MyISAM DEFAULT CHARSET=utf8"); 
    if (!$table) { 
die('Could not create table ' . mysqli_error($con)); 
} 
else { 
echo "Table created<br />"; 
} 
$insert = mysqli_query ($con,"INSERT INTO info (city, country) VALUES ('$a', '$b')"); 
if (!$insert) { 
die('Could not insert ' . mysqli_error($con)); 
} 
else { 
echo "Inserted<br />"; 
} 
$select = mysqli_query ($con,"SELECT * FROM info"); 
$res=mysqli_fetch_array($select); 
if (! $res) { 
die('Result not working ' . mysqli_error($con)); 
} 
else { 
echo "Result working<br />"; 
} 

echo "result: ".$res['city']. " "; 
echo "result: ".$res['country']. " "; 
mysqli_close($con); 

Answer 2

表「db_1.info」不存在

意味着，表info不以dB爲單位db_1存在，因此，要確保，如果是這樣的話。

Answer 3

你是不是在$insert上做$table一個mysqli_query()您mysqli_query()面前，你是不是做一個mysqli_query()上$select

$table = "CREATE TABLE info (id INT NOT NULL AUTO_INCREMENT, city CHAR(40), country CHAR(40))"; 
if (! $table) 

$insert = "INSERT INTO info (city, country) VALUES ($a, $b)"; 
if (! $insert) { 

$select = "SELECT * FROM info"; 

$result = mysqli_query ($con, $insert); 
if (! $result) 

嘗試添加mysqli_query() -

$table_sql = "CREATE TABLE `info` (`id` INT NOT NULL AUTO_INCREMENT, `city` CHAR(40), `country` CHAR(40), PRIMARY KEY (`id`))"; 
$table = mysqli_query ($con, $table_sql); 
if (! $table) { 
die('Could not create table ' . mysqli_error($con)); 
} 
else { 
echo "Table created<br />"; 
} 

$insert_sql = "INSERT INTO `info` (`city`, `country`) VALUES ('$a', '$b')"; 
$insert = mysqli_query ($con, $insert_sql); 
if (! $insert) { 
die('Could not insert ' . mysqli_error($con)); 
} 
else { 
echo "Inserted<br />"; 
} 

$select = "SELECT * FROM `info`"; 

$result = mysqli_query ($con, $select); 
if (! $result) { 
die('Result not working ' . mysqli_error($con)); 
} 
else { 
echo "Result working<br />"; 
} 

編輯 此外，這條線將失敗 -

echo "result: ".$result['city']. " "; 

，你必須使用mysqli_fetch_array()

$results = mysqli_fetch_array($result); 
echo "result: ".$results['city']. " ";