错误:键"PRIMARY"的条目"0"重复 [英] Error: Duplicate entry '0' for key 'PRIMARY'

问题描述

我无法解决问题，这是我得到的mysql错误:

当数据库中有一条记录时，我可以编辑和更新数据，但是当我添加两行时，就会出现错误.

数据库中的一些图片

当我更改行时，行ID降为0，这是一个问题，因为我无法编辑其他行.

CREATE TABLE `dati` (
 `ID` int(11) NOT NULL AUTO_INCREMENT,
 `title` varchar(255) NOT NULL,
 `value1` varchar(255) NOT NULL,
 `value2` varchar(255) NOT NULL,
 PRIMARY KEY (`ID`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1 PACK_KEYS=1

更新代码:

<?php // Izlabot datus datubāzē!
$titletxt = $_POST['title_edit'];
$value1 = $_POST['value1_edit'];
$value2 = $_POST['value2_edit'];

if(isset($_POST['edit'])){
$con=mysqli_connect("localhost","root","","dbname");
if (mysqli_connect_errno())
  {
  echo "Neizdevās savienoties ar MySQL: " . mysqli_connect_error();
  }
$sql="UPDATE dati SET ID='$ID',title= '$titletxt',value1='$value1',value2='$value2' WHERE 1";
if (!mysqli_query($con,$sql))
  {
  die('Error: ' . mysqli_error($con));
  }
echo '<script>
        alert(" Ieraksts ir veiksmīgi labots! ");

        window.location.href = "index.php";
    </script>';
mysqli_close($con);
}
?>

从表单:

<?php
            $con=mysqli_connect("localhost","root","","dbname");
            if (mysqli_connect_errno())
              {
              echo "Neizdevās savienoties ar MySQL: " . mysqli_connect_error();
              }
            $result = mysqli_query($con,"SELECT * FROM dati");
            while($row = mysqli_fetch_array($result))
              {
              echo "<tr>";
                  echo "<td><input id='titled' type='text' name='title_edit' value='" . $row['title'] . "'></td>";
                  echo "<td><input id='value1d' type='text' name='value1_edit' value='" . $row['value1'] . "'></td>";
                  echo "<td><input id='value2d' type='text' name='value2_edit' value='" . $row['value2'] . "'></td>";
                  echo "<input type='hidden' name='id' value='" . $row['ID'] . "'>";
                  echo "<td><button name='edit' id='edit_btn' class='frm_btns' value='" . $row['ID'] . "'>Edit</button></td>";
              echo "</tr>";
              }
            mysqli_close($con);
        ?>

由于返回0，因此无法读取ID的值.

解决方案

对于那些由于问题标题而到达此问题的人(如我所做的那样)，这解决了我的问题:

此错误可能表明表的PRIMARY KEY未设置为AUTO-INCREMENT，(并且您的插入查询未指定ID值).

要解决:

检查您的表上是否设置了主键，并且已将主键设置为自动插入".

如何使用phpmyadmin将自动增量添加到mysql数据库中的列?

I can't resolve my problem, this is the error from mysql that I'm getting:

I can edit and update my data when I've got one record in the database but when I add two rows, I get the error.

Some pictures from database

And when I change the row, row ID goes down to 0 and that's is a problem as I can't edit other rows.

CREATE TABLE `dati` (
 `ID` int(11) NOT NULL AUTO_INCREMENT,
 `title` varchar(255) NOT NULL,
 `value1` varchar(255) NOT NULL,
 `value2` varchar(255) NOT NULL,
 PRIMARY KEY (`ID`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1 PACK_KEYS=1

Update Code:

<?php // Izlabot datus datubāzē!
$titletxt = $_POST['title_edit'];
$value1 = $_POST['value1_edit'];
$value2 = $_POST['value2_edit'];

if(isset($_POST['edit'])){
$con=mysqli_connect("localhost","root","","dbname");
if (mysqli_connect_errno())
  {
  echo "Neizdevās savienoties ar MySQL: " . mysqli_connect_error();
  }
$sql="UPDATE dati SET ID='$ID',title= '$titletxt',value1='$value1',value2='$value2' WHERE 1";
if (!mysqli_query($con,$sql))
  {
  die('Error: ' . mysqli_error($con));
  }
echo '<script>
        alert(" Ieraksts ir veiksmīgi labots! ");

        window.location.href = "index.php";
    </script>';
mysqli_close($con);
}
?>

From form:

<?php
            $con=mysqli_connect("localhost","root","","dbname");
            if (mysqli_connect_errno())
              {
              echo "Neizdevās savienoties ar MySQL: " . mysqli_connect_error();
              }
            $result = mysqli_query($con,"SELECT * FROM dati");
            while($row = mysqli_fetch_array($result))
              {
              echo "<tr>";
                  echo "<td><input id='titled' type='text' name='title_edit' value='" . $row['title'] . "'></td>";
                  echo "<td><input id='value1d' type='text' name='value1_edit' value='" . $row['value1'] . "'></td>";
                  echo "<td><input id='value2d' type='text' name='value2_edit' value='" . $row['value2'] . "'></td>";
                  echo "<input type='hidden' name='id' value='" . $row['ID'] . "'>";
                  echo "<td><button name='edit' id='edit_btn' class='frm_btns' value='" . $row['ID'] . "'>Edit</button></td>";
              echo "</tr>";
              }
            mysqli_close($con);
        ?>

It couldn't read the value of ID, as 0 was returned.

解决方案

For those arriving at this question because of the question title (as I did), this solved my problem:

This error can indicate that the table's PRIMARY KEY is not set to AUTO-INCREMENT, (and your insert query did not specify an ID value).

To resolve:

Check that there is a PRIMARY KEY set on your table, and that the PRIMARY KEY is set to AUTO-INCREMENT.

How to add auto-increment to column in mysql database using phpmyadmin?

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