PHP-MySQL查询拒绝执行

Question

我有以下代码：

<?php

$usStates = array(
    'Alabama',
    'Alaska',
    'Arizona',
    'Arkansas',
    'California',
    'Colorado',
    'Connecticut',
    'Delaware',
    'Florida',
    'Georgia',
    'Hawaii',
    'Idaho',
    'Illinois',
    'Indiana',
    'Iowa',
    'Kansas',
    'Kentucky',
    'Louisiana',
    'Maine',
    'Maryland',
    'Massachusetts',
    'Michigan',
    'Minnesota',
    'Mississippi',
    'Missouri',
    'Montana',
    'Nebraska',
    'Nevada',
    'New Hampshire',
    'New Jersey',
    'New Mexico',
    'New York',
    'North Carolina',
    'North Dakota',
    'Ohio',
    'Oklahoma',
    'Oregon',
    'Pennsylvania',
    'Rhode Island',
    'South Carolina',
    'South Dakota',
    'Tennessee',
    'Texas',
    'Utah',
    'Vermont',
    'Virginia',
    'Washington',
    'West Virginia',
    'Wisconsin',
    'Wyoming',
);

$link = new mysqli("localhost", "drup197", "*******", "census");
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}


foreach ($usStates as $state) {
    printf("Entering loop\n");
    printf("Current state = %s\n", $state);

    if (mysqli_query($link, "CREATE TABLE '$state'_age_and_gender_demographics LIKE national_age_and_gender_demographics")) {
        printf("%s_age_and_gender_demographics successfully created\n", $state);
    }
    if (mysqli_query($link, "CREATE TABLE '$state'_race_demographics_-_one_race LIKE national_race_demographics_-_one_race")) {
        printf("%s_race_demographics_-_one_race successfully craeted\n");
    }
    if (mysqli_query($link, "CREATE TABLE '$state'_race_demographics_-_multi-race LIKE national_race_demographics_-_multi-race")) {
        printf("%s_race_demographics_-_multi-race successfully craeted\n");
    }
    if (mysqli_query($link, "CREATE TABLE '$state'_race_demographics_-_hispanic_or_latino LIKE national_race_demographics_-_hispanic_or_latino")) {
        printf("%s_race_demographics_-_hispanic_or_latino successfully craeted\n");
    }
}
printf("Tables successfuly created");

mysqli_close($link);
?>

基本上，我正在尝试使用状态名称作为表名称的通配符来创建所有50个状态的表。 但是，这些查询似乎没有在执行。 你能告诉我问题吗？

Answer 1

从查询字符串中删除单引号。 您正在尝试创建名为“ California” _age_and_gender_demographics的表。

或尝试：

if (mysqli_query($link, "CREATE TABLE ". $state ."_age_and_gender_demographics LIKE national_age_and_gender_demographics")) {
    printf("%s_age_and_gender_demographics successfully created\n", $state);
}
if (mysqli_query($link, "CREATE TABLE ". $state ."_race_demographics_-_one_race LIKE national_race_demographics_-_one_race")) {
    printf("%s_race_demographics_-_one_race successfully craeted\n");
}
if (mysqli_query($link, "CREATE TABLE ". $state ."_race_demographics_-_multi-race LIKE national_race_demographics_-_multi-race")) {
    printf("%s_race_demographics_-_multi-race successfully craeted\n");
}
if (mysqli_query($link, "CREATE TABLE ". $state ."_race_demographics_-_hispanic_or_latino LIKE national_race_demographics_-_hispanic_or_latino")) {
    printf("%s_race_demographics_-_hispanic_or_latino successfully craeted\n");
}