[英]Inserting a row into a table using PHP MYSQL
以下php代碼用於刪除表(如果存在),創建表,使用表,然后在表中插入一行。
一切都與插入分開。 我是PHP和MYSQL的新手,我嘗試了許多不同類型引號的排列(單引號,雙引號,這是一個單引號),但是無法將數據插入表中。
任何人都可以闡明這是怎么回事嗎?
$retval = mysqli_query($conn,'INSERT INTO `performance` (manager, program, programid, yearmonth, performance) VALUES ("manager1", "program1","programid1", "199901", "-3.4")');
下面的php腳本給出了輸出:
連接成功
表已成功刪除。
數據庫使用成功。
表創建成功。
無法插入數據。
因此,除了插入之外,其他所有東西都起作用。
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully<br />';
$retval = mysql_query('DROP TABLE IF EXISTS `managedfutures`.`performance`') or die(mysql_error());
if(! $retval )
{
die('Could not drop table ' . mysql_error());
}
echo "Table dropped successfully.";
echo "<br>";
$retval = mysql_query("USE managedfutures", $conn);
if(! $retval )
{
die('Could not use DB' . mysql_error());
}
echo "DB used successfully.";
echo "<br>";
$sql = "CREATE TABLE performance( ".
"performance_id INT NOT NULL AUTO_INCREMENT, ".
"manager VARCHAR(255) NOT NULL, ".
"program VARCHAR(255) NOT NULL, ".
"programid VARCHAR(255) NOT NULL, ".
"yearmonth VARCHAR(6) NOT NULL, ".
"performance FLOAT NOT NULL, ".
"PRIMARY KEY (performance_id )); ";
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not create table: ' . mysql_error());
}
echo "Table created successfully.";
echo "<br>";
$retval = mysqli_query($conn,'INSERT INTO `performance` (manager, program, programid, yearmonth, performance) VALUES ("manager1", "program1","programid1", "199901", "-3.4")');
if(! $retval )
{
die('Could not insert data. ' . mysql_error());
}
echo "Data inserted successfully.";
echo "<br>";
return;
感謝Mike W指出我混合了mysql和mysqli命令! 我是php / mysql的新手,但沒有意識到兩者之間存在差異。 還有另一個錯誤,我在插入語句中輸入數字作為字符串。 即我寫的是“ -3.4”而不是-3.4。
為了完整起見,這是可以使用的固定版本。
$mysqli = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
if ($mysqli->connect_errno) {
die("Failed to connect to MySQL: " . $mysqli->connect_error);
}
echo 'Connected successfully<br />';
$retval = mysqli_query($mysqli,"DROP TABLE IF EXISTS `performance`");
if(! $retval )
{
die('Could not drop table ' . $mysqli->query_error);
}
echo "Table dropped successfully.";
echo "<br>";
$sql = "CREATE TABLE performance( ".
"performance_id INT NOT NULL AUTO_INCREMENT, ".
"manager VARCHAR(255) NOT NULL, ".
"program VARCHAR(255) NOT NULL, ".
"programid VARCHAR(255) NOT NULL, ".
"yearmonth VARCHAR(6) NOT NULL, ".
"performance FLOAT NOT NULL, ".
"PRIMARY KEY (performance_id )); ";
$retval = mysqli_query($mysqli, $sql);
if(! $retval )
{
die('Could not create table: ' . $mysqli->query_error);
}
echo "Table created successfully.";
echo "<br>";
$retval = mysqli_query($mysqli, "INSERT INTO `performance` (`manager`, `program`,`programid`, `yearmonth`, `performance`) VALUES ('manager1', 'program1','programid1', '199901', -3.4)");
if(! $retval )
{
die('Could not insert data. ' . $mysqli->query_error);
}
echo "Data inserted successfully.";
echo "<br>";
return;
您正在混合mysql_*()
和mysqli_*()
調用。 兩者是不同的,不能一起使用。 不建議使用mysql_*()
-僅使用mysqli_*()
。
您在整個代碼中都使用了mysql_query
$retval = mysql_query( $sql, $conn ); //**You have used mysql_query**
if(! $retval )
{
die('Could not create table: ' . mysql_error());
}
echo "Table created successfully.";
echo "<br>";
突然看到mysqli_query( MAGIC !!! )。
$retval = mysqli_query($conn,'INSERT INTO `performance` (manager, program, programid, yearmonth, performance) VALUES ("manager1", "program1","programid1", "199901", "-3.4")');
^^
// SUDDENLY you see mysqli_query
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