PHP MYSQL选择并插入但不起作用

Question

I have 2 table: " customer " and " data ". I want to select from " customer " with mysqli_fetch_array() and insert some data to " data " but I can't insert with this code please help me.

<?php
    ini_set('display_errors', 1);
    error_reporting(~0);

    $serverName = "localhost";
    $userName = "root";
    $userPassword = "root";
    $dbName = "mydatabase";

    $conn = mysqli_connect($serverName,$userName,$userPassword,$dbName);

    $sql = "SELECT * FROM customer";

    $query = mysqli_query($conn,$sql);

?>
<table width="600" border="1">
  <tr>
    <th width="91"> <div align="center">URL </div></th>
    <th width="98"> <div align="center">Board </div></th>
    <
  </tr>
<?php
while($result=mysqli_fetch_array($query,MYSQLI_ASSOC))
{
?>
  <tr>
    <td><div align="center"><?php echo $result[$objResult["url"];?></div></td>
    <td><?php echo $result["board"];?></td>

  </tr>
<?php

$sql2 = "INSERT INTO data (url,board) VALUES ('".$result["url"]."','".$result["board"]."')";

    $query2 = mysqli_query($conn,$sql2);


}
?>
</table>
<?php
mysqli_close($conn);
?>

Answer 1

I found 1 syntax error in your code.

You have error here

<?php echo $result[$objResult["url"];?>

which should be

<?php echo $result["url"];?>

check my code.

<?php
    ini_set('display_errors', 1);
    error_reporting(~0);

    $serverName = "localhost";
    $userName = "root";
    $userPassword = "";
    $dbName = "mydatabase";

    $conn = mysqli_connect($serverName,$userName,$userPassword,$dbName);

    $sql = "SELECT * FROM customer";

    $query = mysqli_query($conn,$sql);
?>
<table width="600" border="1">
  <tr>
    <th width="91"> <div align="center">URL </div></th>
    <th width="98"> <div align="center">Board </div></th>
      </tr>
<?php
while($result=mysqli_fetch_array($query,MYSQLI_ASSOC))
{
?>
  <tr>
    <td>
        <div align="center">
        <?php echo $result["url"];?>
        </div>
    </td>
    <td><?php echo $result["board"];?></td>

  </tr>
<?php

$sql2 = "INSERT INTO data (url,board) VALUES ('".$result["url"]."','".$result["board"]."')";
$query2 = mysqli_query($conn,$sql2);
}
?>
</table>
<?php
mysqli_close($conn);
?>

dont forget to change your connection parameters.

here is the SQL Script :

DROP TABLE IF EXISTS customer ; CREATE TABLE customer ( id int(11) NOT NULL AUTO_INCREMENT, url varchar(255) DEFAULT NULL, board varchar(255) DEFAULT NULL, PRIMARY KEY ( id ) ) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1;

---

-- Records of customer

---

INSERT INTO customer VALUES ('1', 'First URL', 'Bord 1'); INSERT INTO customer VALUES ('2', 'Second URL', 'Board 2'); INSERT INTO customer VALUES ('3', null, null);

---

-- Table structure for data

---

DROP TABLE IF EXISTS data ; CREATE TABLE data ( id int(11) NOT NULL AUTO_INCREMENT, url varchar(255) DEFAULT NULL, board varchar(255) DEFAULT NULL, PRIMARY KEY ( id ) ) ENGINE=InnoDB AUTO_INCREMENT=11 DEFAULT CHARSET=latin1;

---

-- Records of data

---

INSERT INTO data VALUES ('1', 'First URL', 'Bord 1'); INSERT INTO data VALUES ('2', 'First URL', 'Bord 1'); INSERT INTO data VALUES ('3', 'Second URL', 'Board 2'); INSERT INTO data VALUES ('4', 'First URL', 'Bord 1'); INSERT INTO data VALUES ('5', 'Second URL', 'Board 2'); INSERT INTO data VALUES ('6', 'First URL', 'Bord 1'); INSERT INTO data VALUES ('7', 'Second URL', 'Board 2'); INSERT INTO data VALUES ('8', 'First URL', 'Bord 1'); INSERT INTO data VALUES ('9', 'Second URL', 'Board 2'); INSERT INTO data VALUES ('10', '', '');