MySQL Update SET替换为PHP多个查询

Question

I'm trying to run 3 lines of the same code as seen below.I've also tried the W3Schools PHP mysqli_multi_query() Function but it gives me a syntax error and only the 1-st line of code works the ExistingWord1

I also read a couple of stackoverflow posts about this and this can't figure it out

<?php
include 'words.php';
include('../config.php');
$servername = DB_HOSTNAME;
$username = DB_USERNAME;
$password = DB_PASSWORD;
$dbname = DB_DATABASE;
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
/*UTF-8 format*/
$conn->set_charset("utf8");

$sql = "UPDATE product_description SET name = REPLACE(name, '$ExistingWord1', '$ReplacerWord1') WHERE name LIKE '%$SearchWord1%';";
$sql = "UPDATE product_description SET name = REPLACE(name, '$ExistingWord2', '$ReplacerWord2') WHERE name LIKE '%$SearchWord2%';";
$sql = "UPDATE product_description SET name = REPLACE(name, '$ExistingWord3', '$ReplacerWord3') WHERE name LIKE '%$SearchWord3%';";

if ($conn->query($sql) === TRUE) {
    echo "Products updated successfully";
} else {
    echo "Error updating record: " . $conn->error;
}

$conn->close();
?>

This is the full Code that i am trying to run as by itself it only runs the 1st $sql =""

and ignores the other 2

as for the variables in the REPLACE

$ExistingWord1 words that are in the database name
$ReplacerWord1  replaces the ExistingWord1
'%$SearchWord1%' real database names

Thanks for the help guys

Everyone in this post was really helpful i really do appriciate it.

Answer 1

According to the fragment provided only last $sql value will be executed, since you redefine same variable 3 times, only the last value stored in var $sql.

To fix that you can try:

$sql = "UPDATE product_description SET name = REPLACE(name, '$ExistingWord1', '$ReplacerWord1') WHERE name LIKE '%$SearchWord1%';";
$sql .= "UPDATE product_description SET name = REPLACE(name, '$ExistingWord2', '$ReplacerWord2') WHERE name LIKE '%$SearchWord2%';";
$sql .= "UPDATE product_description SET name = REPLACE(name, '$ExistingWord3', '$ReplacerWord3') WHERE name LIKE '%$SearchWord3%';";

UPDATE So now you have just debug your query and code any preffered way. For example:

 $sql = "UPDATE product_description SET name = REPLACE(name, '$ExistingWord1', '$ReplacerWord1') WHERE name LIKE '%$SearchWord1%';";
if ($conn->query($sql) === TRUE) {
    echo "1. Products updated successfully";
} else {
    echo "Error updating record: " . $conn->error;
}
    $sql = "UPDATE product_description SET name = REPLACE(name, '$ExistingWord2', '$ReplacerWord2') WHERE name LIKE '%$SearchWord2%';";
if ($conn->query($sql) === TRUE) {
    echo "2. Products updated successfully";
} else {
    echo "Error updating record: " . $conn->error;
}

    $sql = "UPDATE product_description SET name = REPLACE(name, '$ExistingWord3', '$ReplacerWord3') WHERE name LIKE '%$SearchWord3%';";

if ($conn->query($sql) === TRUE) {
    echo "3. Products updated successfully";
} else {
    echo "Error updating record: " . $conn->error;
}