LIKE运算子无法在MYSQL中运作

Question

this is my code when i run it

$name=$_GET['name'];
$strSQL = "SELECT * FROM category WHERE name LIKE =" .$name;

$rs = mysql_query($strSQL);

// Loop the recordset $rs
$response = array();
while($row = mysql_fetch_array($rs)) {


    $product = array();
    $product["id"] = $row["id"];
    $product["name"] = $row["name"];
    array_push($response, $product);
}

echo json_encode($response);
}
error:mysql_fetch_array() expects parameter 1 to be resource, boolean given in while($row = mysql_fetch_array($rs))

it shows error on this line help me to fix this

Answer 1

It is because You are not sending Query matched with your datatype of the name field.

Try below syntax with ''

$strSQL = "SELECT * FROM category WHERE name LIKE '" .$name . "'";

IF you still getting the same error then try this code

$name=$_GET['name'];
$strSQL = "SELECT * FROM category WHERE name LIKE '" .$name . "'";

$rs = mysql_query($strSQL);
if(mysql_num_rows($rs) > 0){
// Loop the recordset $rs
$response = array();
while($row = mysql_fetch_array($rs)) {


    $product = array();
    $product["id"] = $row["id"];
    $product["name"] = $row["name"];
    array_push($response, $product);
}

echo json_encode($response);
}
}
else{ echo json_encode(array('msg'=>"No records found")); }

Answer 2

您插入了错误的语法....使用此

 $strSQL = "SELECT * FROM category WHERE name LIKE '%$name'";

Answer 3

You need to open a connection to your database. Something like:

<?php
$dbhost = 'localhost:3036';
$dbuser = 'root';
$dbpass = 'rootpassword';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}
$name=$_GET['name'];
$strSQL = "SELECT * FROM category WHERE name LIKE '%" .$name . "%'";
// $strSQL = "SELECT * FROM category WHERE name = '" .$name . "'";

$rs = mysql_query($strSQL, $conn);

// Loop the recordset $rs
$response = array();
while($row = mysql_fetch_array($rs)) {


    $product = array();
    $product["id"] = $row["id"];
    $product["name"] = $row["name"];
    array_push($response, $product);
}

echo json_encode($response);
?>

Also if you are going to use "LIKE" in your sql, you don't use "=".