我想合并 MySQL 中的两个表:
table_1
--------------------------------------------------------
id | table_1_name | table_1_art
--------------------------------------------------------
1 | Ernest Hemingway | Writer
--------------------------------------------------------
2 | Mark Twain | Writer
--------------------------------------------------------
3 | Jackson Pollock | Painter
--------------------------------------------------------
4 | Miles Davis | Musician
--------------------------------------------------------
5 | Akira Kurosawa | Cinema
--------------------------------------------------------
table_2
--------------------------------------------------------
id | table_2_name | table_2_art
--------------------------------------------------------
1 | Hemingway, ernest | Writer
--------------------------------------------------------
2 | Twain, Mark | Writer
--------------------------------------------------------
3 | P., Jackson | Painter
--------------------------------------------------------
4 | Miles Dewey Davis | Musician
--------------------------------------------------------
5 | Bukowski, Charles | Poetry
--------------------------------------------------------
我想用 table_1_name 的数据更新字段 table_2_name，但前提是它们之间存在相似性。我正在使用 similar_text，并选择超过 40% 的相似度:
<?php
// Connection
$server = "****";
$user = "****";
$pass = "****";
$db_name = "****";
$conn = new mysqli($server, $user, $pass, $db_name);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} else {
echo "Connected <br/>";
}
//Begin
$petition = "SELECT * FROM table_1, table_2";
$result = mysqli_query($conn, $petition);
while ($row = mysqli_fetch_assoc($result)) {
$table_1_name = $row['table_1_name'];
$table_1_art = $row['table_1_art'];
$table_2_name = $row['table_2_name'];
$table_2_art =$row['table_2_art'];
//Ask for similarities
similar_text($table_1_name, $table_2_name, $percent);
//Show coincidences over 40%
if ($percent > "40%") {
echo $table_1_name." — ".$table_2_name." — ".$percent."% <br />";
//Update field in table_2_name for this coincidences
$sql = "UPDATE table_2 SET table_2_name = '$table_1_name'";
if ($conn->query($sql) === true) {
echo "updated <br/> <br/>";
} else {
echo "ERROR: " . $conn->error." <br/> <br/>";
}
}
$conn->close();
}
但是我得不到结果。而不是这个:
table_2
--------------------------------------------------------
id | table_2_name | table_2_art
--------------------------------------------------------
1 | Ernest Hemingway | Writer
--------------------------------------------------------
2 | Mark Twain | Writer
--------------------------------------------------------
3 | Jackson Pollock | Painter
--------------------------------------------------------
4 | Miles Davis | Musician
--------------------------------------------------------
5 | Bukowski, Charles | Poetry
--------------------------------------------------------
我明白了:
table_2
--------------------------------------------------------
id | table_2_name | table_2_art
--------------------------------------------------------
1 | Ernest Hemingway | Writer
--------------------------------------------------------
2 | Ernest Hemingway | Writer
--------------------------------------------------------
3 | Ernest Hemingway | Painter
--------------------------------------------------------
4 | Ernest Hemingway | Musician
--------------------------------------------------------
5 | Ernest Hemingway | Poetry
--------------------------------------------------------
我敢肯定这是一个很常见的错误，至少对于初学者来说是这样，但我找不到!
尼基塔
PD:正如@denny 所说，sql 需要一个条件，我忘了说我尝试了一些选项，例如:
$sql = "UPDATE table_2 SET table_2_name = '$table_1_name' WHERE ($percent > '45%')";
现在的问题是:哪个条件以及如何。不敢相信有那么难……
最佳答案
修改sql
$sql = "UPDATE table_2 SET table_2_name = '$table_1_name'";
为此
$sql = "UPDATE table_2 SET table_2_name = '$table_1_name'"where column="您的条件";
语法:-
更新 table_name set column_name = 'value' where column_name ='your_condition'
如果你执行
更新 table_name set column_name = 'value'
它将更新整个表格行
关于php - 使用 PHP similar_text 和 UPDATE 合并表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42461163/