Question

我试图将数据插入到数据库中，我不知道为什么它不起作用。

/// <binding Build='default' />

var del = require('del'),
    gulp = require("gulp"),
    ts = require('gulp-typescript'),
    watch = require('gulp-watch');

var webproj = "./src/StreakMaker.Web/";
var webroot = webproj + "wwwroot/";
var appsource = webproj + "App/";
var appout = webroot + "app/";
var jspmsource = "./jspm_packages/";
var jspmout = webroot + "jspm_packages/";

var paths = {
    webroot: webroot,
    src: appsource,
    app: appout,
    jspm: jspmsource,
    jspm_out: jspmout
};

gulp.task('watch', ['watch-typescript', 'watch-html']);

gulp.task('watch-typescript', function(){
    gulp.watch(paths.src + '/**/*.ts', ['build-typescript']);
});

gulp.task('clean-typescript', function () {
    del([paths.app + '/**/*.ts']);
});

gulp.task('build-typescript', ['clean-typescript'], function () {
    var tsProject = ts.createProject('./tsconfig.json');

    gulp.src(paths.src + '/**/*.ts')
        .pipe(ts(tsProject))
        .pipe(gulp.dest(paths.app));
});

gulp.task('watch-html', function () {
    gulp.watch(paths.src + '/**/*.html', ['copy-html']);
});

gulp.task('clean-html', function () {
    del([paths.app + '/**/*.html']);
});

gulp.task('copy-html', ['clean-html'], function () {
    gulp.src(paths.src + '/**/*.html')
        .pipe(gulp.dest(paths.app));
});

gulp.task('copy-jspm', ['clean-jspm', 'copy-config'], function() {
    gulp.src(paths.jspm + "**/*.{js,css,map}")
        .pipe(gulp.dest(paths.jspm_out));
});

gulp.task('clean-jspm', function(){
    del([paths.jspm_out + "**/*.*"]); 
});

gulp.task('copy-config', ['clean-config'], function(){
    gulp.src("./config.js")
        .pipe(gulp.dest(paths.webroot));
});

gulp.task('clean-config', function(){
    del(paths.webroot + 'config.js'); 
});

gulp.task('default', ['build-typescript', 'copy-html', 'copy-jspm']);

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Answer 1

您应该构建查询PHP样式，而不是C样式。即使这样，我也不建议mysql，mysqli更好。另外，请查看PDO以获得更高级别的安全性。

//设置查询

$q = "INSERT INTO `pm` ('sent_to', 'sent_by', 'date', 'title', 'content', 'status')) VALUES(
$MSGInfo['sent_to'],$MSGInfo['sent_by'],$date,
mysql_real_escape_string($_POST['reply_title']), mysql_real_escape_string($_POST['reply_conent']),Unread)";

//运行查询

$result = mysql_query($q) or die(mysql_error());

你的查询构建得不是很好，你需要多做一些工作。

将数据插入数据库不起作用（PHP）

1 个答案: