PHP insert语句仅插入1次

Question

我使用php向 table.ive插入变量$email、$starcount、$bookid固定值，现在只是为了测试文件。$res查询检查其中是否有带有该图书id和电子邮件的行。如果没有$sql查询，则$nex查询循环使用任何starcount列，其中book列=$bookE 219。

如果我更改文件顶部的电子邮件，它应该插入到新的信息数据库中，并取出新的和现有的星号，但它不会发布，它只是返回已经存在的星号。我不明白为什么它不起作用..。我使用数组返回到我的文件。

<?php
mysql_connect("localhost","root","");
mysql_select_db("FunReads");

$email = "sd";
$starcount = "2";
$bookid = "5";

$res = mysql_query("SELECT * FROM Review WHERE book_id='$bookid' AND user_email='$email'");

if (mysql_num_rows($res) != 0) {

    $array[]= array("starcount" => "already entered");

} else {

    $sql = mysql_query("INSERT INTO Review(book_id,starcount,user_email) values('.$bookid.','.$starcount.','.$email')");

    $nex = mysql_query("SELECT * FROM Review WHERE book_id='$bookid'");

    while($row = mysql_fetch_array($nex)){

        $star = $row["starcount"];

        $array[] = array("starcount" => $star);

    }
}

echo json_encode($array); 
//echo "[{"name":"n1","city":"NY"},{"name":"n2","city":"Paris"}, ...]

?>

Answer 1

在我看来，"Review“表中的"book_id”是主键，因为您多次尝试添加它，所以系统会显示错误重复键。检查错误并张贴。还请检查insert查询是否工作。

Answer 2

您不应该手动传递主键值，它将帮助您。

<?php
 mysql_connect("localhost","root","");
 mysql_select_db("FunReads");
 $starcount="2";
 $email = "vinodh@gmail.com";
  $res=mysql_query("SELECT * FROM Review  WHERE email   ='$email'");
  if(mysql_num_rows($res)!=0){
      $array[]= array("starcount" => "already entered");
  }else{
         $sql=mysql_query("INSERT INTO Review (starcount,email)    values('.$starcount.','.$email')");
$nex=mysql_query("SELECT * FROM Review  WHERE email='$email'");
    while($row=mysql_fetch_array($nex)){
            $star = $row["starcount"];
            $array[] = array("starcount" => $star);
    }
  }
  echo json_encode($array);
 ?>

Answer 3

我刚刚更新了你的代码，它对我很好。

​

<?php
mysql_connect("localhost","user","");
mysql_select_db("xyz");

$email = "hari@gmail.com";
$starcount = "2";
$bookid = "5";

$sql = "SELECT * FROM review WHERE book_id='$bookid' AND user_email='$email'";
$res = mysql_query($sql);

if (mysql_num_rows($res) != 0) {

    $array[]= array("starcount" => "already entered");

} else {
	$sql = "INSERT INTO review(book_id,starcount,user_email) values('$bookid','$starcount','$email')";
    $sql = mysql_query($sql);

    $nex = mysql_query("SELECT * FROM review WHERE book_id='$bookid'");

    while($row = mysql_fetch_array($nex)){

        $star = $row["starcount"];

        $array[] = array("starcount" => $star);

    }
}

echo json_encode($array); 

​

示例输出:

{“星号”：“2”}，{“星号”：“3”}，{“星号”：“1”}，{“星号”：“2”}，{“星号”：“1”}

我更新了insert查询，请尝试更新并测试。