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พอดีผมส่งค่าไปอีกหน้าเพื่อโชว์ข้อมูลครับ ขึ้น Error Warning: implode(): Invalid arguments passed in C:\xampp\htdocs\siamleathergoods\test_product1_test.php on line 4 ยังไงรบกวนชี้แนะแนวทางด้วยครับ ขอบคุณครับ
หน้าส่งค่าครับ
<?php
$con = mysqli_connect("localhost", "root", "", "ecommshoes");
$get_pro_style = "select * from product where product_cat = 1 LIMIT 0,6 ";
$run_pro_style = mysqli_query($con, $get_pro_style);
while($row_pro_style = mysqli_fetch_array($run_pro_style)){
?>
<form action="test_product1_test.php" method="post" name="form1">
<label>
<input type='radio' name='shoes_style' value='<?php echo $row_pro_style["product_id"];?>'>
<?php echo $row_pro_style["product_id"];?><br />
<?php echo $row_pro_style["product_title"];?><br />
<img src='admin_area/product_image/<?php echo $row_pro_style["product_image"];?>' /><br />
------------------------------------------------<br />
</label>
<?php
}
?>
<input type="submit" value="Provide Fit" name="submit" id="submit" class="button-button" />
</form>
หน้าแสดงค่าที่ส่งมาครับ
<?php
error_reporting(E_ALL & ~E_NOTICE);
$con = mysqli_connect("localhost", "root", "", "ecommshoes");
$in_value = implode("," , $_POST["shoes_style"]);
$sql = mysqli_query($con, "select * from product where product_cat in ('$in_value')");
?>
<?php
while( $r = mysqli_fetch_assoc($sql)){
?>
<?php echo $r["product_id"];?><br />
<?php echo $r["product_title"];?><br />
<?php echo $r["product_price"];?><br />
<img src='admin_area/product_image/<?php echo $r["product_image"];?>' /><br />
<?php
}
?>
Tag : PHP, MySQL
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Date :
2017-01-11 17:53:35 |
By :
Junior.J |
View :
3756 |
Reply :
3 |
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