Question

我在尝试将用户变量插入数据库时遇到了一些严重困难。我一直收到错误您的SQL语法错误。我已经搜索了几天的表格并尝试了不同的解决方案，但无济于事。我尝试使用'$var'，'".$var."'，"$var"但没有运气。如果我使用没有引号的$ var我得到错误列数与第1行的值计数不匹配。任何帮助将不胜感激。谢谢。

下面是代码段和数据表。

$con= mysqli_connect("oni", "mul", "kjnbiunvtr","mul"); 

if (mysqli_connect_errno($con))
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$drugName=$_POST['drugName']; 
$drugName=mysql_real_escape_string($drugName);

$treats=$_POST['treats'];
$treats=mysql_real_escape_string($treats);

$effects=$_POST['effects']; 
$effects =mysql_real_escape_string($effects);

$sideEffect=$_POST['SideEffects']; 
$sideEffect= mysql_real_escape_string($sideEffect);

$sideEffect2=$_POST['SideEffects2']; 
$sideEffect2=mysql_real_escape_string($sideEffect2);

$sideEffect3=$_POST['SideEffects3']; 
$sideEffect3= mysql_real_escape_string($sideEffect3);

$sql=mysqli_query($con,"INSERT INTO Drug(drug) VALUES('$drugName')");

if (mysqli_query($con,$sql))
{
    echo "Drug insertion successful : ";
    echo "\r\n";
}
else
{
    echo "Error 1 :  " . mysqli_error($con);
    echo "\r\n";
}

表：

$sql1="CREATE TABLE Drug
    (Id int (11) NOT NULL AUTO_INCREMENT,
        PRIMARY KEY(Id),
        drug VARCHAR(50) NOT NULL
    )ENGINE=INNODB;";


$sql2="CREATE TABLE DrugSideEffect
    ( Id int (11) NOT NULL AUTO_INCREMENT,
        D_Id int (11) NOT NULL,
        SE_Id int (11) NOT NULL,
        PRIMARY KEY(Id),
        FOREIGN KEY (SE_Id) REFERENCES SideEffect(Id),
        FOREIGN KEY (D_Id) REFERENCES Drug(Id)
        ON DELETE CASCADE
        ON UPDATE CASCADE
    )ENGINE=INNODB;";


$sql3="CREATE TABLE Treatment
    (Id INT NOT NULL AUTO_INCREMENT,
        T_Id int (11) NOT NULL,
        Descr VARCHAR (250) NOT NULL,
        PRIMARY KEY(Id),
        FOREIGN KEY (T_Id) REFERENCES Drug(Id)
        ON DELETE CASCADE
        ON UPDATE CASCADE
    )ENGINE=INNODB;";


$sql4="CREATE TABLE Effects
    (Id INT NOT NULL AUTO_INCREMENT,
        E_Id int (11) NOT NULL,
        Descr VARCHAR (250) NOT NULL,
        PRIMARY KEY(Id),
        FOREIGN KEY (E_Id) REFERENCES Drug(Id)
        ON DELETE CASCADE
        ON UPDATE CASCADE
    )ENGINE=INNODB;";


$sql5="CREATE TABLE SideEffect
    (Id int (11) NOT NULL AUTO_INCREMENT,
        PRIMARY KEY(Id),
        Descr VARCHAR (250) NOT NULL
    )ENGINE=INNODB";

Answer 1

也许你的MySQL版本是＆gt; = 5.5并且不推荐mysql_real_escape_string函数。您可以尝试使用mysqli_real_escape_string方法，如下例所示。

$con= mysqli_connect("oni", "mul", "kjnbiunvtr","mul"); 

if (mysqli_connect_errno($con))
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$drugName=$_POST['drugName']; 
$drugName=mysqli_real_escape_string($con, $drugName);

$sql=mysqli_query($con,"INSERT INTO Drug(drug) VALUES('$drugName')");

if (mysqli_query($con,$sql))
{
    echo "Drug insertion successful : ";
    echo "\r\n";
}
else
{
    echo "Error 1 :  " . mysqli_error($con);
    echo "\r\n";
}

Answer 2

试试这个

     $sql=mysqli_query($con,"INSERT INTO Drug (drug) VALUES ('".$drugName."')");

编辑：

你的问题在于创建表格。

您的表格排序不佳，您应该在SideEffect之前创建表格DrugSideEffect。

尝试这个顺序：

$sql1="CREATE TABLE Drug
(Id int (11) NOT NULL AUTO_INCREMENT,
    PRIMARY KEY(Id),
    drug VARCHAR(50) NOT NULL
)ENGINE=INNODB;";
 $sql5="CREATE TABLE SideEffect
(Id int (11) NOT NULL AUTO_INCREMENT,
    PRIMARY KEY(Id),
    Descr VARCHAR (250) NOT NULL
)ENGINE=INNODB";

$sql2="CREATE TABLE DrugSideEffect
( Id int (11) NOT NULL AUTO_INCREMENT,
    D_Id int (11) NOT NULL,
    SE_Id int (11) NOT NULL,
    PRIMARY KEY(Id),
    FOREIGN KEY (SE_Id) REFERENCES SideEffect(Id),
    FOREIGN KEY (D_Id) REFERENCES Drug(Id)
    ON DELETE CASCADE
    ON UPDATE CASCADE
)ENGINE=INNODB;";


$sql3="CREATE TABLE Treatment
(Id INT NOT NULL AUTO_INCREMENT,
    T_Id int (11) NOT NULL,
    Descr VARCHAR (250) NOT NULL,
    PRIMARY KEY(Id),
    FOREIGN KEY (T_Id) REFERENCES Drug(Id)
    ON DELETE CASCADE
    ON UPDATE CASCADE
)ENGINE=INNODB;";


$sql4="CREATE TABLE Effects
(Id INT NOT NULL AUTO_INCREMENT,
    E_Id int (11) NOT NULL,
    Descr VARCHAR (250) NOT NULL,
    PRIMARY KEY(Id),
    FOREIGN KEY (E_Id) REFERENCES Drug(Id)
    ON DELETE CASCADE
    ON UPDATE CASCADE
)ENGINE=INNODB;";

Answer 3

$con= mysqli_connect("oni", "mul", "kjnbiunvtr","mul");

if (mysqli_connect_errno($con)) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); }

$drugName=$_POST['drugName'];

$drugName=mysqli_real_escape_string($con, $drugName);

$sql=mysqli_query($con,"INSERT INTO Drug(drug) VALUES('$drugName')");

if（mysqli_query（$ con，$ sql））{ echo "Drug insertion successful : "; echo "\r\n"; } else { echo "Error 1 : " . mysqli_error($con); echo "\r\n"; }

PHP SQL /您的SQL语法有错误

3 个答案: